Assignment 1. Prolog to Problem 1. Two cylinders. ü Visualization. Problems by Branko Curgus

Transcription

1 Assignment In[]:= Problems by Branko Curgus SetOptions $FrontEndSession, Magnification Prolog to Problem. Two cylinders In[]:= This is a tribute to a problem that I was assigned as an undergraduate student in the mid-970s at the Department of Mathematics of the University of Sarajevo. The problem involves the intersection of two circular cylinders. The smaller cylinder is vertical with the equation: x + y = x (this is the circle centered at 0 with the radius ) and z is arbitrary. The larger cylinder is horizontal with the equation: x + z = and y is arbitrary. I do not recall exactly what I was asked to do with these two cylinders. However, I remember well that I spent a lot of time trying to visualize the intersection. At that time I did not have access to any technology to help with visualization. Since I do not recall what was the exact question related to these two cylinders I will calculate all quantities that come to mind related to this object. ü Visualization This is a convenient view point in the graph below which I determined through experimentation. vp.3`,.`,.` Out[]=.3,.,.

2 070_A.nb In[]:= cyl ParametricPlot3D Cos t, Cos t Sin t z t,, z, 3, 3 PlotStyle Opacity.8 Mesh None, PlotPoints 00, 00 ; cyl ParametricPlot3D Cos t y, Sin t y, 3.3, 3.3 t, 0, PlotStyle Opacity.3 Mesh 3, 6 ; int ParametricPlot3D Cos t, Cos t Sin t Cos t t,, PlotStyle Thickness Blue ; int ParametricPlot3D Cos t, Cos t Sin t Cos t t,, PlotStyle Thickness Blue ; cirh ParametricPlot3D Cos t, Cos t Sin t 0 t,, PlotStyle Thickness Red, Opacity Mesh None ; cirv ParametricPlot3D Cos t 0,Sin t t,, PlotStyle Thickness Red, Opacity Mesh None ; CylCyl Show cyl, cyl, int, int, cirh,cirv, PlotRange.,.6.6,.6 5.3, 5.3 AxesEdge AxesLabel x, y, z ImageSize 00, ViewPoint vp

3 Out[8]= 070_A.nb 3

4 070_A.nb In[9]:= sur ParametricPlot3D s Cos t, Cos t Sin t Cos t s Cos t, Cos t Sin t Cos t t, 0, s, 0, PlotStyle Opacity.9 Mesh 0, 0 PlotPoints 50, 50 ; sur ParametricPlot3D s Cos t, Cos t Sin t Cos t s Cos t, Cos t Sin t Cos t t, 0, s, 0, PlotStyle Opacity.9 Mesh None, PlotPoints 50, 50 ; sur3 ParametricPlot3D s Cos t, Cos t Sin t Cos t s Cos t, Cos t Sin t Cos t t,, s, 0, PlotStyle Opacity.9 Mesh 0, 0 PlotPoints 00, 00 ; int ParametricPlot3D Cos t, Cos t Sin t Cos t t,, PlotStyle Thickness Blue ; int ParametricPlot3D Cos t, Cos t Sin t Cos t t,, PlotStyle Thickness Blue ; cir ParametricPlot3D Cos t, Cos t Sin t 0 t,, PlotStyle Thickness Red, Opacity Mesh None ;

5 070_A.nb 5 In[]:= CylCyl Show cyl, sur, sur, sur3, int, int, PlotRange.,.6.6,.6 5.3, 5.3 AxesEdge AxesLabel x, y, z ImageSize 00, ViewPoint vp Out[]=

6 6 070_A.nb In[5]:= CylCyl3 Show sur, sur, sur3, int, int, PlotRange.,.6.6,.6 5.3, 5.3 AxesEdge AxesLabel x, y, z ImageSize 00, ViewPoint vp Out[5]=

7 070_A.nb 7 In[6]:= surp ParametricPlot3D s Cos t, Cos t Sin t Cos t s Cos t, Cos t Sin t Cos t.6,0,0 t, 0, s, 0, PlotStyle Opacity.9 Mesh 0, 0 PlotPoints 50, 50 ; surp ParametricPlot3D s Cos t, Cos t Sin t Cos t s Cos t, Cos t Sin t Cos t.6,0,0 t, 0, s, 0, PlotStyle Opacity.9 Mesh None, PlotPoints 50, 50 ; sur3p ParametricPlot3D s Cos t, Cos t Sin t Cos t s Cos t, Cos t Sin t Cos t 0, 0 t,, s, 0, PlotStyle Opacity.9 Mesh 0, 0 PlotPoints 00, 00 ; int ParametricPlot3D Cos t, Cos t Sin t Cos t t,, PlotStyle Thickness Blue, Opacity Mesh None ; int ParametricPlot3D Cos t, Cos t Sin t Cos t t,, PlotStyle Thickness Blue, Opacity Mesh None ; CylCyl Show surp, surp, sur3p, int, int, PlotRange.,.6.6,.6 5.3, 5.3 AxesEdge AxesLabel x, y, z ImageSize 00, ViewPoint vp

8 8 070_A.nb Out[0]= ü Unwrapping Each cylinder can be unwrapped to a plane. Hence the blue curve above and two surfaces bounded by the blue curve can be unwrapped into planes. How to find the equation for the blue curve when unwrapped to a plane? The following picture can help.

9 070_A.nb 9 In[]:= Show surp, surp, sur3p, ParametricPlot3D Cos Cos Cos, Sin, PlotStyle Blue, Thickness 0.0 ParametricPlot3D Cos Cos Cos, Sin, PlotStyle Blue, Thickness 0.0 cirh, cirv, PlotRange.,.6.6,.6 5.3, 5.3 AxesEdge AxesLabel x, y, z ImageSize 00, ViewPoint vp Out[]= ü Large cylinder Let us consider first unwrapping the blue curve into the plane by unwrapping the larger cylinder. As the larger cylinder is unwrapped to a plane, the vertical red half-circle is unwrapped to a vertical line segment of length p. The parametric equation of the top part of the blue curve is Cos t, Cos t Sin t Cos t The equation of the red horizontal circle is Cos t, Cos t Sin t 0 and the blue curve is above the red horizontal circle and on the large cylinder whose top part is given by z = - x. Now we will determine the equation of the top part of the blue curve in relation to the arc length on the vertical red circle. Let P be the point on the vertical red circle such that s is the arc length between the point 0, 0. The

10 0 070_A.nb coordinates of P are Cos s 0, Sin s. Which point on the blue curve has the same z-coordinate? Call that point B. The point B has the same x-coordinate as P. Thus, Cos s Cos t Solving for t we get In[]:= Solve Cos s Cos t,t Solve::ifun : Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information. à Out[]= t ArcCos Cos s t ArcCos Cos s t ArcCos Cos s t ArcCos Cos s Substituting t ArcCos Cos s in Cos t, Cos t Sin t Cos t In[3]:= FullSimplify Cos t, Cos t Sin t Cos t. t ArcCos Cos s And 0 s, s Out[3]= Cos s Sin s Csc s Sin s, Sin s This expression can be further simplified to In[]:= Cos s Sin s Csc s Sin s Cos s, Sin s Out[]= Cos s Cos s Sin s Sin s Thus, the blue point B is at the distance Sin s Cos s from the point P. Thus, when flattened the equation of the blue curve is as follows:

11 070_A.nb In[5]:= Plot Cos s Sin s Cos s Sin s s, PlotStyle Thickness Blue Thickness Blue AspectRatio Automatic, Ticks Range, Range PlotRange.,..,. ImageSize 00 Out[5]= -p - 3 p - p - p p p 3 p p - It is interesting to point out that the equivalent expression for the blue curve above is as follows In[6]:= FullSimplify Cos s Cos s Cos s Sin s And 0 s, s Out[6]= True ü Small cylinder Now we will unwrap the blue curve by unwrapping the smaller cylinder. As a parameter we will take the arc length s along the horizontal red circle starting from 0, 0. As before, call P the point on this circle such that the arc length between P and 0, 0 is s. From the geometry of the unit circle it follows the polar angle q for this point is q =s. Therefore this point is given as Cos s, Cos s Sin s 0 The parametric equation for the top part of the blue curve is Cos t, Cos t Sin t Cos t Thus the point on the blue curve vertically above the point P is Cos s, Cos s Sin s Cos s.

12 070_A.nb In[7]:= Plot Cos s, Cos s s, PlotStyle Thickness Blue Thickness Blue AspectRatio Automatic, Ticks Range, Range PlotRange.,..,. ImageSize 00 Out[7]= -p - 3 p - p - p p p 3 p p - - Or, cutting the surface at the point 0, 0. In[8]:= Plot Cos, Cos PlotStyle Thickness Blue Thickness Blue AspectRatio Automatic, Ticks Range, Range PlotRange.,..,. ImageSize 00 Out[8]= -p - 3 p - p - p p p 3 p p - - ü Calculations ü The surface area of the eight-shaped region This is the surface area on the larger cylinder. It is the transparent surface in the picture. The area is

13 070_A.nb 3 In[9]:= TrigToExp Integrate Cos s Sin s s, 0, Out[9]= 6 6 Log In[30]:= N 6 6 Log Out[30]= 8.55 A different way to calculate this surface is to use the surface integral. In[3]:= Cross D x, y, x x D x, y, x y Out[3]= x x,0, In[3]:= Simplify x x,0,. x x,0,,x 0 Out[3]= x In[33]:= Solve x y x,y Out[33]= y x x y x x In[3]:= Integrate x, y, x x, x x Out[3]= x x x In[35]:= TrigToExp Integrate Integrate x, y, x x, x x x, 0, Out[35]= 6 6 Log In[36]:= N Out[36]= 8.55 Exactly the same value. ü The vertical surface area We found the equation so we just integrate. In[37]:= TrigToExp Integrate Cos s, s, 0, Out[37]= 8 Log 8 Log In[38]:= N 8 Log 8 Log Out[38]= 8.367

14 070_A.nb This expression can be simplified by hand to In[39]:= 8 Log 3 Out[39]= 8 Log 3 In[0]:= N 8 Log 3 Out[0]= It is interesting to point out that the above formula for the blue curve has equivalent form as follows: In[]:= FullSimplify Cos s Cos s Out[]= True In[]:= TrigToExp Integrate Cos t, t, 0, Out[]= 8 Log Log In[3]:= N 8 Log Log Out[3]= A different way to calculate this surface is to use the surface integral. I was not able to get the exact value in this way, just an approximation. To get equation for the surface In[]:= Solve x y, y Out[]= y x x y x x In[5]:= FullSimplify Cross D x, z, x x x D x, z, x x z Out[5]= x x x,0, In[6]:= Simplify x x x,0,. x x x,0,,x 0 Out[6]= x x We need to integrate this quantity over the half circle in xz-plane In[7]:= Integrate Integrate, x,0, y y, x x But this integral is not evaluating. Try for the indefinite integral, that is anti-derivative:

15 070_A.nb 5 In[8]:= FullSimplify Integrate x x,x And x 0, x Out[8]= Log x x Now use the Fundamental Theorem of Calculus In[9]:= FullSimplify Log x x. x y Log x x. x 0 And y x Out[9]= Log Log y y And integrate numerically: In[50]:= Chop NIntegrate Log Log y y y, Out[50]= ü The volume To calculate the volume we integrate the big cylinder over the small disk: In[5]:= TrigToExp Integrate Integrate x, y, x x, x x x, 0, Out[5]= 3 Log In[5]:= N 3 Out[5]= Log Or the small cylinder over the big half-circle. I was not able to get the exact value in this way. In[53]:= TrigToExp Integrate Integrate x x, x,0, z z, In[5]:= Integrate x x, x, 0, z Assumptions And z 0, z Out[5]= z z z z ArcSin In[55]:= NIntegrate z z z z ArcSin z, 0, Out[55]= ü The length of the blue curve: We first use the first parametrization in the plane obtained above:

16 6 070_A.nb In[56]:= FullSimplify D Cos s Sin s s Out[56]= 3 Cos s Sec s In[57]:= NIntegrate 3 Cos s Sec s, s, 0, Out[57]= The second parametrization in the plane In[58]:= FullSimplify D Cos s,s Out[58]= 9 6 Cos s 3 Cos s In[59]:= NIntegrate Out[59]= Cos s 3 Cos s, s, 0, Finally we use the parametrization of the blue curve in space: In[60]:= Cos t, Cos t Sin t Cos t Out[60]= Cos t, Cos t Sin t Cos t In[6]:= FullSimplify D Cos t, Cos t Sin t Cos t t Out[6]= In[6]:= Cos t Sin t Cos t Cos t 3 Sin t Cos t FullSimplify Cos t Sin t Cos t Cos t 3 Sin t Cos t. Cos t Sin t Cos t Cos t 3 Sin t Cos t 6 Out[6]= 9 Cos t 3 Cos t 6 In[63]:= NIntegrate 9 Cos t, t,, 3 Cos t Out[63]= One more parametrization in space

17 070_A.nb 7 In[6]:= D Cos s Cos s Sin s Sin s s Out[6]= s Sin s Cos Cos s Sin s Sin s, Cos s Cos s In[65]:= s FullSimplify Sin s Cos Cos s Sin s Sin s, Cos s. Cos s s Sin s Cos Cos s Sin s Sin s, Cos s Cos s Out[65]= In[66]:= 3s Cos Sec s NIntegrate 3s Cos Sec s, s, 0, Out[66]= Problem : Your task In[67]:= Repeat as much as you can of what I did in Prolog to Problem when the large Cylinder is replaced by the sphere of radius centered at the origin vp `, `, ` Out[67]=.386,.0909,.8707

18 8 070_A.nb In[68]:= cyl ParametricPlot3D Cos t, Cos t Sin t z t,, z, 3, 3 PlotStyle Opacity Mesh None, PlotPoints 00, 00 ; sph ParametricPlot3D Cos Sin Sin Sin Cos 0, 0, PlotStyle Opacity.6 Mesh 8, ; CylSph Show cyl, sph, PlotRange.,.6.6,.6 5.3, 5.3 AxesEdge AxesLabel x, y, z ImageSize 00, ViewPoint vp Out[70]= Prolog to Problem. Two famous multiple integrals In[7]:= ü A famous double integral Consider the integral 0 0 x y. - x y Mathematica does calculate this integral. Integrate Integrate, x, 0, y, 0, x y Out[7]= 8 However, calculating this integral by hand using the Fundamental theorem of calculus is not easy. Here is why

19 070_A.nb 9 In[7]:= Out[7]= In[73]:= Out[73]= In[7]:= Out[7]= In[75]:= Out[75]= Integrate x y,x ArcTanh x y y This is doable by hand. Now we apply the FTC and get ArcTanh x y y. x ArcTanh y y Next we integrate Integrate ArcTanh y,y y PolyLog y PolyLog y ArcTanh x y. x 0 y But this integral is a special function that is not familiar to an undergraduate student. In fact this is not often studied function. Applying FTC again we get. PolyLog y PolyLog y. y 8 PolyLog y PolyLog y. y 0 Our goal is to understand this result using change of variables in double integral. We use the following change of variables: x u v uv, y u v uv Good way to think about this change of variables is as a mapping from uv-plane into xy-plane. u, v is being mapped in u v uv, u v uv One way to visualize this mapping is to plot the grid of coordinate lines in uv-plane and map them into curves in xyplane. For example the line u=/ is being mapped into the curve { v v, v v }

20 0 070_A.nb In[76]:= GraphicsGrid Graphics Table Hue u Line u, 5 u, 5 u,,.5 PlotRange Frame True, ImageSize 5, FrameLabel "u", "v" RotateLabel False Graphics Table Hue u PointSize Table Point u v uv, u v v, 5, 5, u,,.5 vu 00 PlotRange Frame True, ImageSize 5, FrameLabel "x", "y" RotateLabel False Out[76]= v 0 y u x

21 070_A.nb In[77]:= GraphicsGrid Graphics Table Hue v Line 5, v 5, v v,,.5 PlotRange Frame True, ImageSize 5, FrameLabel "u", "v" RotateLabel False Graphics Table Hue v PointSize Table Point u v uv, u v u, 5, 5, v,,.5 vu 00 PlotRange Frame True, ImageSize 5, FrameLabel "x", "y" RotateLabel False Out[77]= v 0 y u x Since we see some overlap it is clear that we will have to narrow down the area in uv-plane.

22 070_A.nb In[78]:= GraphicsGrid Graphics Table Hue u Thickness 0.0 Line u, u, u, 0.975,,.05 PlotRange Frame True, ImageSize 5, FrameLabel "u", "v" RotateLabel False Graphics Table Hue u Thickness 0.0 Line Table u v uv, u v v,, u,.975,,.05 vu 00 PlotRange Frame True, ImageSize 5, FrameLabel "x", "y" RotateLabel False Out[78]= v 0.0 y u x

23 070_A.nb 3 In[79]:= GraphicsGrid Graphics Table Hue v 3 Thickness 0.0 Line v v v, 0.975,,.05 PlotRange Frame True, ImageSize 5, FrameLabel "u", "v" RotateLabel False Graphics Table Hue v 3 Thickness 0.0 Line Table u v uv, u v u,, v, 0.975,,.05 vu 00 PlotRange Frame True, ImageSize 5, FrameLabel "x", "y" RotateLabel False Out[79]= v 0.0 y u x

24 070_A.nb In[80]:= GraphicsGrid Graphics Table Hue u Thickness Line u, u, u, 0.975,,.05 Table Hue v 3 Thickness Opacity 0.5 Line v v v, 0.975,,.05 PlotRange Frame True, ImageSize 5, FrameLabel "u", "v" RotateLabel False Graphics Table Hue u Thickness Line Table u v uv, u v v,, u,.975,,.05 vu 00 Table Hue v 3 Thickness Opacity 0.5 Line Table u v uv, u v u,, v, 0.975,,.05 vu 00 PlotRange Frame True, ImageSize 5, FrameLabel "x", "y" RotateLabel False

25 070_A.nb 5 Out[80]= Just for fun

26 6 070_A.nb In[8]:= Manipulate GraphicsGrid Graphics Table Hue t Point rr Cos t Sin t t, 0,, 6 PlotRange Frame True, ImageSize 5, FrameLabel "u", "v" RotateLabel False Graphics Table Hue t Point rr Cos t Sin t rr Cos t Sin t Cos t Sin t rr Sin t Cos t t, 0,, 6 PlotRange Frame True, ImageSize 5, FrameLabel "x", "y" RotateLabel False rr,.5.,.9 rr Out[8]= v 0.0 y u x

27 070_A.nb 7 In[8]:= Manipulate GraphicsGrid Graphics Table Hue k Thickness 0.0 Line rr Cos k Sin k rr Cos k Sin k k, 0, 8, 6 6 PlotRange Frame True, ImageSize 5, FrameLabel "u", "v" RotateLabel False Graphics Table Hue t Thickness 0.0 Line rr Cos t Sin t rr Cos t Sin t Cos t Sin t rr Sin t Cos t rr Cos t Sin t 6 6 rr Cos t Sin t, Cos t Sin t 6 rr Sin t Cos t t, 0, PlotRange Frame True, ImageSize 5, FrameLabel "x", "y" RotateLabel False rr,.5.0,.99

28 8 070_A.nb rr Out[8]= v 0.0 y u x In[83]:= Out[83]= In[8]:= Out[8]= In[85]:= Out[85]= Let us go back to the change of variables: x u v uv, y u v uv This is how Mathematica calculates the Jacobian matrix: D u v u v, u v u, v u v v u v u u v, uv uv u v v uv uv u u v, uv uv uv uv We need the determinant of the Jacobian matrix FullSimplify Det D u v u v, u v u, v u v u v u v We notice that this quantity is negative for u, v being in [-,]. So the absolute value is FullSimplify Det D u v u v, u v u, v u v u v u v We need to understand which set in uv-plane is being mapped onto the unit square with vertices (0,0), (,0), (,), (,0) in xy-plane. Next we list points in uv-plane and their images in xy-plane (0,0) goes to (0,0) (,) goes to (,0)

29 070_A.nb 9 (,-) goes to (0,) (,0) goes to (,) This indicates that the triangle with vertices (0,0), (,-), (,) is being mapped onto the unit square. The following graphics confirms that. In[86]:= GraphicsGrid Graphics Table Hue u Line Table u, v v, u, u,.0 u, 0,.00,.05 PlotRange 0, Frame True, ImageSize 00, FrameLabel "u", "v" RotateLabel False Graphics Table Hue u Line Table u v uv, u v v, u, u,.0 u, 0,.00,.05 vu PlotRange 0, 0, Frame True, ImageSize 5, FrameLabel "x", "y" RotateLabel False Out[86]= v 0.0 y x u

30 30 070_A.nb In[87]:= Out[87]= In[88]:= Now we substitute the new variables into the integrand FullSimplify u v u v u v u v x y. x uv,y vu We notice that this quantity is positive. This expression needs to be multiplied by the absolute value of the determinant of the Jacobian matrix. Thus, the new integrand is u v FullSimplify u v Det D u v uv, u v u, v uv Out[88]= u v So, we need to calculate the integral 0 u u v u. u v In[89]:= This integral is easily calculable by hand. However, for efficiency I do it in Mathematica here. Integrate, v, u, u u v Out[89]= In[90]:= Out[90]= In[9]:= ArcTan u u Now we integrate with respect to u. This is also a simple change of variables integral. Integrate ArcTan u, u, 0, u 8 ü A famous triple integral We consider the integral x y z. + x yz Mathematica does calculate this integral. Integrate Integrate Integrate, x, 0, y, 0, z, 0, x y z Out[9]= 3 3 Our goal is to understand this result using change of variables in double integral. The change of variables here is x Sin u Cos v, y Sin v Cos w, z Sin w Cos u Good way to think about this change of variables is as a mapping from uvw-space into xyz-space.

31 070_A.nb 3 In[9]:= u, v, w is being mapped in Sin u Cos v, Sin v Cos w, Sin w Cos u This is how Mathematica calculates the Jacobian matrix: D Sin u Cos v, Sin v Cos w, Sin w u, v, w Cos u Out[9]= Cos u Sec v Sec v Sin u Tan v 0 0, Cos v Sec w Sec w Sin v Tan w Sec u Sin w Tan u 0,Cos w Sec u In[93]:= We need the determinant of the Jacobian matrix FullSimplify Det D Sin u Cos v, Sin v Cos w, Sin w u, v, w Cos u Out[93]= Tan u Tan v Tan w This function is always positive. So, no problems here. We need to understand which set in uv-plane is being mapped onto the unit cube with the vertices (0,0,0), (,0,0), (,,0), (,0,0), (0,0,), (,0,), (,,), (,0,) in xyz-space. Next we list points in uvw-space and their images in xyz-space (0,0,0) goes to (0,0,0) ( p,0,0) goes to (,0,0) (0, p,0) goes to (0,,0) (0,0, p ) goes to (0,0,) ( p, p, p ) goes to (,,) This indicates that the polytope with vertices (0,0,0), ( p,0,0), (0, p,0), (0,0, p,0), ( p, p, p ) is being mapped onto the unit cube. Our next goal is to confirm that. Here is the image of the polytope:

32 3 070_A.nb In[9]:= Graphics3D Blue, Thickness Line 0, 0, 0,0,0 Line 0, 0, 0 0,,0 Line 0, 0, 0 0, 0, Line,0,0 0,,0 0, 0,,0,0 Opacity 0.6 Polygon 0, 0, 0 0, 0, 0,,0 Polygon 0, 0, 0,0,0 0,,0 Polygon 0, 0, 0,,0 Polygon,0,0 0,,0 Opacity. Polygon 0, 0, 0,0,0 0, 0, Polygon 0, 0,,0,0 PlotRange 0, 0, 0, Ticks Range 0,, Range 0,, Range 0,, Boxed True, Axes True, AxesLabel x, y, z ImageSize 00 Out[9]=

33 070_A.nb 33 This polytope can be described mathematically as x, y, z œ R 3 x > 0, y > 0, z > 0, x + y < p, y + z < p, z + x < p In[95]:= Mathematica can plot this polytope using these inequalities: RegionPlot3D And x 0, y 0, z 0, x y,y z,x z x, 0, y, 0, z, 0, PlotPoints 65, 65, 65 PlotStyle Opacity.5 Mesh False, AxesLabel x, y, z Ticks Range 0,, Range 0,, Range 0,, Out[95]= Next we will use Manipulate to slice this polytope into triangles which we will map into the unit cube.

34 3 070_A.nb In[96]:= Manipulate Graphics3D Blue, Thickness Line 0, 0, 0,0,0 Line 0, 0, 0 0,,0 Line 0, 0, 0 0, 0, Line,0,0 0,,0 0, 0,,0,0 Opacity 0.6 Polygon 0, 0, 0 0, 0, 0,,0 Polygon 0, 0, 0,0,0 0,,0 Polygon 0, 0, 0,,0 Polygon,0,0 0,,0 Opacity. Polygon 0, 0, 0,0,0 0, 0, Polygon 0, 0,,0,0 Opacity If a, Polygon a,0,0 0, a,0 0, 0, a Polygon a,0,0 a a 0,,0 a a 0, 0, a PlotRange 0, 0, 0, Ticks Range 0,, Range 0,, Range 0,, Boxed True, Axes True, ImageSize 00 a,.5 0,

35 070_A.nb 35 a Out[96]= In[97]:= Out[97]= In the above Manipulation I used Polygon command to draw the slicing triangles. For that I needed only the vertices of the triangles. However, to map the triangles into xyz-space I need to represent the triangles using two parameters s and t. How to do that? If we have two points say A and B, then all the point on the line segment AB are given by - s A + sb where 0 s. We use the same idea for three points, say A, B, C. First we represent all the points on the line segments AB and AC - s A + sb, 0 s and - s A + sc, 0 s. Now for a fixed s I use parameter t to represent all the point on the line segment connecting the point - s A + sb and - s A + sc : - t -s A + sb + t - s A + sc where 0 s, t. Simplifying the last expression we get the following formula for the surface of the triangle with vertices A, B, C Clear pa, pb, pc, s, t ; Collect FullSimplify t s pa spb t s pa spc pa, pb, pc pa s pc s t pb s st

36 36 070_A.nb Thus the formula - s A + s - t B + stc where 0 s, t represents the surface of the triangle with vertices A, B, C. I will illustrate this with a graphics: In[98]:= Manipulate If a, Show Graphics3D Blue, Thickness Line 0, 0, 0,0,0 Line 0, 0, 0 0,,0 Line 0, 0, 0 0, 0, Line,0,0 0,,0 0, 0,,0,0 Opacity 0.6 Polygon 0, 0, 0 0, 0, 0,,0 Polygon 0, 0, 0,0,0 0,,0 Polygon 0, 0, 0,,0 Polygon,0,0 0,,0 Opacity. Polygon 0, 0, 0,0,0 0, 0, Polygon 0, 0,,0,0 ParametricPlot3D s a,0,0 t s 0, a,0 s t 0, 0, a s, 0, t, 0, PlotRange 0, 0, 0, Ticks Range 0,, Range 0,, Range 0,, Boxed True, Axes True, ImageSize 300 Show Graphics3D Blue, Thickness Line 0, 0, 0,0,0 Line 0, 0, 0 0,,0 Line 0, 0, 0 0, 0, Line,0,0 0,,0 0, 0,,0,0 Opacity 0.6 Polygon 0, 0, 0 0, 0, 0,,0 Polygon 0, 0, 0,0,0 0,,0 Polygon 0, 0, 0,,0 Polygon,0,0 0,,0

37 070_A.nb 37 Opacity. Polygon 0, 0, 0,0,0 0, 0, Polygon 0, 0,,0,0 ParametricPlot3D s t s s t a 0, a 0, 0, a,0,0 a,0 a a, s, 0, t, 0, PlotRange 0, 0, 0, Ticks Range 0,, Range 0,, Range 0,, Boxed True, Axes True, ImageSize 300 a,.5 0, a Out[98]= Next I use the mapping u, v, w is being mapped in Sin u Cos v, Sin v Cos w, Sin w Cos u

38 38 070_A.nb In[99]:= Manipulate If a, ParametricPlot3D Evaluate Sin u Cos v, Sin v Cos w, Sin w Cos u. u a s v a s t w a st, s, 0, t, 0, PlotRange 0, 0, 0, Ticks Range 0,,. Range 0,,. Range 0,,. Boxed True, Axes True, ImageSize 300 ParametricPlot3D Evaluate Sin u Cos v, Sin v Cos w, Sin w. u 3 a a s Cos u v a s as a st w a a st, s, 0, t, 0, PlotRange 0, 0, 0, Ticks Range 0,,. Range 0,,. Range 0,,. Boxed True, Axes True, ImageSize 300 a,.5 0, a Out[99]= In[00]:= Manipulate If a, GraphicsGrid Show Graphics3D Blue, Thickness Line 0, 0, 0,0,0

39 070_A.nb 39 Line 0, 0, 0 0,,0 Line 0, 0, 0 0, 0, Line,0,0 0,,0 0, 0,,0,0 Opacity 0.6 Polygon 0, 0, 0 0, 0, 0,,0 Polygon 0, 0, 0,0,0 0,,0 Polygon 0, 0, 0,,0 Polygon,0,0 0,,0 Opacity. Polygon 0, 0, 0,0,0 0, 0, Polygon 0, 0,,0,0 ParametricPlot3D s a,0,0 t s 0, a,0 s t 0, 0, a s, 0, t, 0, PlotRange 0, 0, 0, Ticks Range 0,, Range 0,, Range 0,, Boxed True, Axes True, ImageSize 600, ParametricPlot3D Evaluate Sin u Cos v, Sin v Cos w, Sin w Cos u. u a s v a s t w a st, s, 0, t, 0, PlotRange 0, 0, 0, Ticks Range 0,,. Range 0,,. Range 0,,. Boxed True, Axes True, ImageSize 600 ImageSize 500 GraphicsGrid Show Graphics3D Blue, Thickness Line 0, 0, 0,0,0 Line 0, 0, 0 0,,0 Line 0, 0, 0 0, 0, Line,0,0 0,,0 0, 0,,0,0 Opacity 0.6 Polygon 0, 0, 0 0, 0, 0,,0 Polygon 0, 0, 0,0,0 0,,0 Polygon 0, 0, 0,,0 Polygon,0,0 0,,0

40 0 070_A.nb Opacity. Polygon 0, 0, 0,0,0 0, 0, Polygon 0, 0,,0,0 ParametricPlot3D s t s s t a 0, a 0, 0, a,0,0 a,0 a a, s, 0, t, 0, PlotRange 0, 0, 0, Ticks Range 0,, Range 0,, Range 0,, Boxed True, Axes True, ImageSize 600, ParametricPlot3D Evaluate Sin u Cos v, Sin v Cos w, Sin w. u 3 a a s Cos u v a s as a st w a a st, s, 0, t, 0, PlotRange 0, 0, 0, Ticks Range 0,,. Range 0,,. Range 0,,. Boxed True, Axes True, ImageSize 600 ImageSize 500 a,.5 0, a Out[00]=

41 070_A.nb In[0]:= Out[0]= In[0]:= Out[0]= Now we substitute the new variables into the integrand FullSimplify Tan u Tan v Tan w Sin u Sin v. x,y x y z Cos v Cos w,z Sin w Cos u This expression needs to be multiplied by the absolute value of the determinant of the Jacobian matrix. Thus, the new integrand is FullSimplify Tan u Tan v Tan w Det D Sin u Cos v, Sin v Cos w, Sin w u, v, w Cos u Thus, the integral is simply the volume of the polytope that found. To calculate the volume of the polytope we split it into two pyramids. One pyramid has vertices 0, 0, 0 p,0,0 0, p,0 0, p,0. The base of this pyramid is equilateral triangle with the vertices p,0,0 0, p,0 0, p,0 and the height of this pyramid is the line segment with the endpoints 0, 0, 0 p, p, p. Thus the volume of this pyramid is *base*height = 3 * p 3 * p 8 6 * 3 = p3 8 One pyramid has vertices 0, 0, 0 p,0,0 0, p,0 0, p,0 p, p, p. The base of this pyramid is equilateral triangle with the vertices p,0,0 0, p,0 0, p,0 and the height of this pyramid is the line segment with the endpoints p, p, p p, p, p. Thus the volume of this pyramid is *base*height = 3 * p 3 * p 8 * 3 = p3 96 Thus the volume of the polytope is p p3 96 = p3 3 Problem. Your task Calculate the famous double integral by using a change of variables similar to what was used to calculate the famous triple integral. Provide all the illustrations as I did in my calculation of the famous double integral. Problem 3. Cissoid of Diocles ü Prolog to Cissoid of Diocles problem This problem is about the curve called Cissoid of Diocles. In fact this is a family of curves. Below is the construction of the Cissoid of Diocles based on circle with radius one.

42 070_A.nb In[03]:= Manipulate Graphics Thickness 0.00 RGBColor.75,.75,.75 Line 0, 0 Cos Sin Thickness RGBColor 0, 0,.6 Line 9, 9, Thickness RGBColor 0, 0,.6 Circle 0, Thickness 0.00 RGBColor 0.8,0,0 Line Table Cos t Sin t Cos t Sin t Sin t t, 0.0,.0,.0 Sin t Thickness RGBColor, 0 Line Sin Cos Sin Sin Cos Sin Thickness 0.00 RGBColor 0, 0, 0 Line 0, 0 Cos Sin Cos Sin Sin Sin PointSize 0.0 RGBColor 0, 0,.6 Point Cos Sin PointSize RGBColor 0.8,0,0 Table Point Cos t Sin t Cos t Sin t Sin t t, Sin t PointSize 0.0 RGBColor 0.8,0,0, 8 8 Point Cos Sin Cos Sin Sin Sin PointSize 0.0 RGBColor 0.8,0,0 Point 0, 0 PointSize 0.0 RGBColor 0, 0,.6 Point Sin Cos Sin Sin PlotRange 5, 5 0,.3 Frame False, Axes True, Ticks Range 6, 6 Range 0, 3 ImageSize 500, 8 8 q Out[03]=

43 070_A.nb 3 ü Your task Learn about the Cissoid of Diocles family of curves. The corresponding Wikipedia page is very informative. Based on that information answer the following two questions: ü Length Calculate the length of the part of the Cissoid of Diocles which is inside the circle which is used in the definition of that Cissoid. That is the maroon length in the picture below. In[0]:= Graphics Thickness RGBColor 0.8,0,0 Line Table Cos t Sin t Cos t Sin t Sin t t,, 3, 6 Sin t Thickness RGBColor 0, 0,.6 Circle 0, PlotRange.,. 0.,. Frame False, Axes False, ImageSize 00 Out[0]= In your response to this question you should present and illustrate all the relevant information that you found on the Internet. You should present at least two distinct calculations for the length. The exact value for the length is expected. ü Area Calculate the area which is below the Cissoid of Diocles and which is inside the circle which is used in the definition of that Cissoid. That is the red area in the picture below.

44 070_A.nb In[05]:= Graphics RGBColor 0., 0. Polygon Join Table Cos t Sin t Cos t Sin t Sin t t,, Sin t 3 6 Table Sin Cos Sin Sin 3, 5 6 Thickness RGBColor 0.8,0,0 Line Table Cos t Sin t Sin t Cos t Sin t Sin t t,, 3, 6 Thickness RGBColor 0, 0,.6 Circle 0, PlotRange.,. 0.,. Frame False, Axes False, ImageSize 00 Out[05]= In your response to this question you should present and illustrate all the relevant information that you found on the Internet. You should present at least two distinct calculations for the area. The exact value for the area is expected. Problem. Explore Ellipses Let a and b be positive real numbers. Consider the ellipse x a y b. The parametric equation for this ellipse is

45 070_A.nb 5 a Cos t b Sin t 0 t. In this problem we consider only the ellipses in this canonical form. ü Area ü Calculate Present at least three different ways to evaluate the area enclosed by this ellipse. ü Together The most famous circle is the unit circle. The area enclosed by the unit circle is p. Determine the family of all ellipses each one of which encloses the area p. Replicate the picture below which shows 0 such ellipses ü Length ü Calculate Determine the exact formula for the circumference of the ellipse introduced in the introduction to this problem. Determine at least three different, but equivalent, ways of writing this formula. Give explanations of the functions that you are using. ü Together The most famous circle is the unit circle. The the circumference of the unit circle is p. Determine the family of all ellipses each one of which has the circumference p. Replicate the picture below which shows 0 such ellipses.

46 6 070_A.nb p - p - p p p - p - p - p

47 070_A.nb 7 p - p - p p p - p - p - p