Lesson 10. Transforming 3D Integrals

Transcription

1 Lesson 10 Transforming D Integrals

2 Example 1: Triple Integrals to Compute Volume ecall that in previous chapters we could find the length of an interval I by computing dx or the area of a region by computing I It follows that we can compute the volume of a -dimensional region by calculating 1 dv. For this cube, the calculation is not exciting: dx dy dz x y 4z z z1 x1 x1 y y0 dy dz dz da. 8 1 x 1, 0 y, 1 z

3 Example : Triple Integrals to Compute Volume We can spice things up a bit. We can find the volume of the region above the xy-plane, the xz-plane, the yz-plane, and below the plane given by x y 4z 8 : 8 x y If x y 4z 8, then z. So z-top is the 4 8 x y plane z and z-bottom is the plane z x y Find the intersection between z and 4 z 0 : We get the line y 8 x, which is our y-top. Our y-bottom is the line y 0. Finally, integrate x from x = 0 to x = 8. y 8xy 8 8x dz dy dx x

4 Example : Triple Integrals to Compute Volume We can spice things up a bit. We can find the volume of the region above the xy-plane, the xz-plane, the yz-plane, and below the plane given by x y 4z 8 : 8xy 8 8x 4 8 8x 1 1 dz dy dx (8 x y) dy dx y8x 1 y 8y xy dx 4 0 y0 8(8 x) x(8 x) (8 x) dx 1 V A h pyramid base

5 Example : Changing the Order of Integration epeat Example by computing dy dx dz where is the region above the xy-plane, the xz-plane, the yz-plane, and below the plane given by x y 4z 8 If x y 4z 8, then y 8 x 4z. So y-top is 8x4z and y-bottom is the plan e y 0. x y 4z 8 and y 0 intersect at the line x 8 4z (our x-top). Our x-bottom is the line x 0. Finally, integrate z from z = 84z 8x4z dy dx dz 64 0 to z =. x z

6 Example 4: D-Change of Variables Compute the volume of the parallelepiped described by the region between the planes z x and z x, y x and y x 4, and y x and y x. First, pick a clever change of variables: u z x, and let u run from 0 to v y x, and let v run from 0 to 4 w y x, and let w run from 0 to Our integral is much more manageable now: dx dy dz V (u,v,w) du dv dw V uvw (u, v,w) du dv dw V (u, v,w) x y z u u u x y z v v v x y z w w w

7 The Volume Conversion Factor: 1 dx dy dz V (u,v,w) du dv dw uvw Let T(u, v, w) be a transformation from uvw-space to -space. That is, T(u,v,w) T (u, v,w),t (u, v,w),t (u, v,w) (x(u, v,w), y(u, v,w),z(u, v,w)). 1 Then V (u, v, w) T v T T 1 1 v T 1 T u u u T T T v T w w w Or V (u, v,w) x y z u u u x y z v v v x y z w w w Note: Just like A (u, v), we need V (u, v, w) to be positive. Hence, the absolute value bars in the formula above. xy

8 Example 4: D-Change of Variables Compute the volume of the parallelepiped described by the region between the planes z x and z x, y x and y x 4, and y x and y x. w v x u z x 1 v w V (u, v,w) 1 v y x y 1 1 w y x 1 z u w v V (u,v,w) du dv dw du dv dw Use.

9 Example 4: D-Change of Variables Compute the volume of the parallelepiped described by the region between the planes z x and z x, y x and y x 4, and y x and y x. No need to study this, it's just nice to verify this works... Check : For a parallelepiped generated by three intersecting vectors X, Y and Z, V (X Y) Z parallelepiped 4 8 Parallelepiped is generated by vectors (1,1,),,, 4, and (0, 0,). V (1,1,) 4, 8, 4 (0, 0,) 8 parallelepiped

10 Example 5: A Mathematica-Assisted Change of Variables Find the volume of the "football" whose outer skin is described by the parametric equation F(s, t) ( 0,0, 4s) 4 s cos(t), sin(t),0 for s and 0 t. First, fill in the football: F(r,s,t) ( 0, 0, 4s) r 4 s cos(t), sin(t), 0 s 0 t 0 r 1 Change of variables: x(r,s, t) r 4 s cos(t) y(r,s, t) r 4 s sin(t) z(r, s,t) 4s V (r,s,t) dr ds dt

11 Example 5: A Mathematica-Assisted Change of Variables Find the volume of the "football" whose outer skin is described by the parametric equation F(s, t ) ( 0,0, 4s) (4 s ) cos(t),sin s and 0 t. x y z V r r r x y z (r,s,t) s s s x y z t t t 4 4(16r 8rs rs ) (Mathematica) (t),0 for V (r,s,t) dr ds dt (Mathematica)

12 Example 5: A Mathematica-Assisted Change of Variables Find the volume of the "football" whose outer skin is described by the parametric equation F(s, t ) ( 0,0, 4s) (4 s ) cos(t),sin s and 0 t. (t),0 for Check using solids of revolution: 8 x 4 dx

13 It is easy to see that harder to interpret Example 6: Beyond Volume Calculations dv computes the volume of a solid. But it's f(x, y,z) dv since f(x,y,z) lives in 4 dimensions. We need an example to keep referring back to to give us some A cube of varying density has its density at each point (x, y,z) described by 4 g f(x, y,z) x y. Find the mass of the cube x y 4 dx dy dz cm 18 grams 15 intuition. 1 x 1, 0 y, 1 z

14 D Integrals: f(x,y,z) dx dy dz f(x(u,v,w),y(u,v,w),z(u,v,w)) V (u,v,w) du dv dw uvw V (u, v, w) x y z u u u x y z v v v x y z w w w In my opinion, a good way to think about as a calculation of the mass of of the solid at any given point (x, y,z). f(x, y,z) dx dy dz is where f(x, y,z) is the density

15 Example 7: A D-Change of Variables with an Integrand Compute 9y dx dy dz where is the parallelepiped that is between the planes z x and z x, y x and y x 4, and y x and y x. w v x All from Example 4: u z x 0 u v w v y x y 0 v 4 w y x 0 w z u w v V (u, v,w) 1 9y dx dy dz y(u, v,w) V (u, v,w) du dv dw v w 1 9 du dv dw 1 v w du dv dw 000

16 Example 8: A Traditional Triple Integral Set up, but do not compute is the paraboloid z 9 x y such that z 0. y dx dy dz where y dz dy dx 9x 0 9x 9x y y dz dy dx z-top is the paraboloid is the plane z 0. z 9 x y and z-bottom Find the intersection between z 0 z 9 x y and : We get the circle x y 9, so our y-top is y 9x and our y-bottom is y 9 x. Finally, integrate x from x to x.