Homework 8. Due: Tuesday, March 31st, 2009

Transcription

1 MATH 55 Applied Honors Calculus III Winter 9 Homework 8 Due: Tuesday, March 3st, 9 Section 6.5, pg. 54: 7, 3. Section 6.6, pg. 58:, 3. Section 6.7, pg. 66: 3, 5, 47. Section 6.8, pg. 73: 33, 38. Section 6.9, pg. 84: 3, 5, 7, 3, 7,. Section 6.5, #8 Consider a square fan blade with sides of length and the lower left corner placed at the origin. If the density of the blade is ρ(x, y) = +.x, is it more difficult to rotate the blade about the x-axis or the y-axis? Solution The lamina in question lies in the (x, y)-plane and is defined by the inequalities x and y. The key quantity of importance in considering resistance to rotation about an axis is the moment of inertia about that axis. The moments of inertia about the x and y-axes are, respectively, I x := y ρ(x, y) da and I y := x ρ(x, y) da, D where D is the square domain occupied by the lamina. Evaluating the moments as iterated integrals, we obtain I x = y ( + x) dx dy [ ] [ ] = y ( dy + x) dx [ y 3 ] y= ( = ) x= y= x x= [ = 8 ( ) 6 5 5] 3 5 = 88 5, D

2 and I y = = [ x 3 = 3 + x4 4 [ 8 = ] 4 [ 8 = 3 + ] 5 = 9 5. x ( + x) dx dy ( x + x3) dx Since I y > I x, it will be harder to rotate the lamina about the y axis than about the x axis. Section 6.6, #4 The figure (in the text) shows the surface created when the cylinder y + z = intersects the cylinder x + z =. Find the area of this surface. Solution It is clear by symmetry that the surface consists of four equal-area faces, so it will be sufficient to pick any one of them and find its surface area, and multiply by four to obtain the final result. Let us find the surface area of one of the yellow-colored faces. This is a surface over the (x, z)-plane. The shadow it casts on the plane is clearly just the cross-section of the cylinder x + z =, that is, the disk x + z. The equation for the surface itself is given by the equation of the intersecting cylinder: y = f(x, z) = z. So, since f x (x, z) = and f z (x, z) = z z, ] x= the surface area of the whole beach-ball will be: S = 4 + fx (x, z) + f z (x, z) da = 4 = 4 x +z x +z x +z x= + z z da da. z Now we have to evaluate this double integral, and I think that the most efficient route to success is to define the disk of integration by the inequalities z and, for each such z, z x z.

3 So, the inner integral will be over x, and the outer integral over z: S = z z z dx dz = z z z dx dz. The inner integral is completely trivial to evaluate, and its value combines nicely with the intermediate factor of / z to make the outer integral just as simple: + + S = 4 z dz = 8 dz = 6. z So the total surface area of the beach ball is S = 6. The double integral could also be evaluated in reverse order with the same result of course, but this results in a lengthier calculation that requires one to remember that arcsin(z) is an antiderivative of / z and also requires a subsequent integration by parts. Another method for solving this problem is to think of the base region for the surface as being the square x and y in the (x, y)-plane. Then the top and bottom of the beach ball (above and below this square in the plane) have equal areas, and the top half of the beach ball may be further divided into four equal-area tiles, a representative of which lies over the triangular region described by the inequalities x and x y. This tile may be further divided into two equal-area parts, a representative of which lies over the region given by x and x y, or equivalently (and more conveniently, it turns out) by y and x y. The height of the upper half of the beach ball above this latter region is given by z = g(x, y) = y, and so we may represent the total surface area of the beach ball as Section 6.7, #5 Evaluate the triple integral S = 6 = 6 = 6 = 6 y y + g x (x, y) + g y (x, y) dx dy dx dy y y y dy = 6 y = 6. d y dy dy E x dv where E is bounded by the paraboloid x = 4y + 4z and the plane x = 4. 3

4 Solution As usual, there are several orders of iterated integration that one could try to use to evaluate this integral, and as usual some of these yield easier calculations than do others. Here is one order that gives a fairly easy answer. We let the outer integration be over x. What is the range of x? Well, the paraboloid x = 4y + 4z opens from the origin toward the positive x-direction, and the plane x = 4 is the boundary in this direction, so the range for x is x 4. Now for each x in this range, the slice through the volume E parallel to the (y, z)-plane is the disk y + z x/4. This disk may be described by the inequalities x y x, and, for each such y, So, we may write the triple integral as an iterated integral as follows: E x dv = 4 + x/ x/ Evaluating the inner integral over z (x is constant) gives E x dv = 4 + x/ x/ x x 4 y dy dx = 4 x x 4 y z 4 y. + x/4 y x dz dy dx. x/4 y xf(x) dx, where f(x) := + x/ x/ x 4 y dy. Now to evaluate f(x), we might try the substitution y = w x/ (recall that x is to be treated as a constant here), which leads to f(x) = x 4 + w dw. The latter integral is just half of the area of the unit circle, so Therefore, E x dv = 4 f(x) = πx 8. xf(x) dx = π 4 4 x dx = π = 6π 3. Another approach is to identify the region E as consisting of the volume between two surfaces x = 4y + 4z and x = 4 over a base that is the disk y + z. Thus, the integral is also easily evaluated by using polar coordinates on the base: E x dv = y +z [ 4 ] x dx da(y, z) = 4y +4z Using polar coordinates to integrate over the disk then yields: E π x dv = 8 ( r 4 ) r dr dθ = 6π y +z ( 4 (4y + 4z ) ) da(y, z). ( r r 5 ) ( dr = 6π ) = 6π

5 Section 6.8, #38 Show that x + y + z e (x +y +z ) dx dy dz = π. (The improper triple integral is defined as the limit of a triple integral over a solid sphere as the radius of the sphere increases indefinitely.) Solution Let I denote the improper triple integral in question. We are told that the meaning of the improper integral is: I := lim R I R, where I R := E(R) x + y + z e (x +y +z ) dv, and E(R) := { (x, y, z) x + y + z R }. Since for each R > the region of integration E(R) is a solid sphere, and since the integrand only involves the distance to the origin, it makes sense to evaluate I R using spherical coordinates. The region E(R) corresponds to the inequalities ρ R, θ π, and φ π in spherical coordinates. Also, x + y + z e (x +y +z ) = ρe ρ and dv = ρ sin(φ) dθ dφ dρ, so I R = R π π Evaluating the inner integral over θ gives ρe ρ ρ sin(φ) dθ dφ dρ. R π π I R = π ρ 3 e ρ sin(φ) dφ dρ because dθ = π. Next, evaluating the next inner integral over φ gives I R = 4π R ρ 3 e ρ dρ because π sin(φ) dφ =. To evaluate the remaining integral requires some kind of integration by parts. Let s proceed this way: first make the substitution u = ρ with ρ dρ = du; therefore R I R = π ue u du. 5

6 Now integrate by parts: R ( ) d I R = π u du e u du = πue u u=r u= + π = πr e R + π R R e u du e u du = πr e R πe u u=r u= = πr e R πe R + π. Taking the limit R everything disappears except for the π: Section 6.9, #7 I = lim R I R = π. (a) Evaluate E dv, where E is the solid enclosed by the ellipsoid x /a + y /b + z /c =. Use the transformation x = au, y = bv, z = cw. (b) The Earth is not a perfect sphere; rotation has resulted in flattening at the poles. So the shape can be approximated by an ellipsoid with a = b = 6378 km and c = 6356 km. Use part (a) to estimate the volume of the Earth. Solution For part (a), we note that the given transformation maps the solid enclosed by the ellipsoid into the solid enclosed by the unit sphere: D = {(u, v, w) u + v + w }. To find out how the volume element dv (x, y, z) is written in terms of the transformed coordinates (u, v, w), we need to calculate the Jacobian: x (x, y, z) u x v x w a (u, v, w) = det y u y v y w = det b = abc. z u z v z w c The formula for the volume element is dv (x, y, z) = (x, y, z) (u, v, w) dv (u, v, w) (note that it is the absolute value of the Jacobian that arises in the formula). Assuming that a, b, and c are all positive, we therefore get dv (x, y, z) = abc dv (u, v, w). The volume contained inside the ellipsoid is therefore V = dv (x, y, z) = (x, y, z) (u, v, w) dv (u, v, w) = abc E D D dv (u, v, w). 6

7 Since the remaining integral is just another way of writing the volume of the unit sphere, we get V = 4 3 πabc. For part (b), we just evaluate this formula for the given values of a, b, and c: Volume of Earth 4 3 π6378 (6356) km 3 =.8 km 3. Section 6.9, # Evaluate the integral by making an appropriate change of variables: ( ) y x cos da y + x where R is the trapezoidal region with vertices (, ), (, ), (, ), and (, ). Solution The region of integration in the (x, y)-plane is shown below: R y R x If we look at the integrand, we see that it involves the combinations of variables u = y x and v = y +x, and moreover, the two diagonal boundaries of the region R correspond to constant values of v, namely v = and v =. So this suggests that we should try to rewrite the integral in terms of the transformation u = y x and v = y + x. Equivalently, we may solve for x and y in terms of u and v: x = v u and y = v + u. The vertical segment x =, y then corresponds to a segment of the line v = u, with u. Also, the horizontal segment y =, x corresponds to a segment of the line v = u, with u. Finally, we noted that the remaining segments of the boundary of R are mapped to segments of the lines v = and v =, so we may assemble this information to come up with a picture of the image S of the trapezoid R in the (u, v)-plane: 7

8 v v = u v = S v = v = u u Now it is easy to see that we could integrate in the (u, v)-plane by taking the outer integral to be over v with v, while the inner integral over u will have limits v u v, and the integrand will take the form cos(u/v). It only remains to write the area element da(x, y) in terms of u and v, so we need the Jacobian [ ] (x, y) (u, v) = det xu x v y u y v = x u y v x v y u = =. The area element is therefore da(x, y) = (x, y) (u, v) da(u, v) = da(u, v). So, as an iterated integral in the (u, v)-plane, we have R cos ( ) y x ( u ) da(x, y) = cos y + x S v da(u, v) = Now, to evaluate the inner integral, we have v v v v ( u ) cos du dv. v ( u ) ( u ) u=v cos du = v sin = v sin() v sin( ) = v sin(). v v u= v (We used the fact that sin( x) = sin(x), that is, sin(x) is an odd function of x.) Substituting this result into the iterated integral leaves just the outer integral: R cos ( ) y x da(x, y) = sin() y + x v dv = sin() v = 3 sin(). It is worth commenting here that in this problem as in problem #7 the Jacobian turned out to be a constant, but in general it is a function of the new variables (u, v, w) or (u, v) whose absolute value must be included in the integrand. 8

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