MA22S1: Problem Set 8 Solutions Due at the end of the tutorial, 2-4 December. 1. Using a double integral, find the volume under the surface z = x2 cos y + x sin 2y , for the region given by R = {(x, y) : 0 ≤ x ≤ 2, 0 ≤ y ≤ π/2}. Solution: We use the region to give us the limits and then the volume is ZZ V = (x2 cos y + x sin 2y)dA R Z π/2 Z 2 dx(x2 cos y + x sin 2y) 0 0 2 Z π/2 1 3 1 2 = dy x cos(y) + x sin(2y) 3 2 0 x=0 Z π/2 8 cos(y) = dy 2 sin(2y) + 3 0 π/2 8 sin(y) = − cos(2y) + 3 y=0 8 sin(π/2) 8 sin(0) = − cos(π) + − − cos(0) + 3 3 14 8 . = 1 + − [−1 + 0] = 3 3 = dy 2. Find the volume below the surface z = e2x , 1 above the region R = {(x, y) : 0 ≤ x ≤ 1, x2 ≤ y ≤ x}. Sketch R. Solution: Let’s first sketch R to understand what region we are integrating over. 1.0 0.8 0.6 x 0.4 x2 0.2 0.0 0.0 0.2 0.4 0.6 0.8 1.0 Figure 1: The region R = {(x, y) : 0 ≤ x ≤ 1, x2 ≤ y ≤ x}. Note that since 0 ≤ x ≤ 1, x2 ≤ x. The only difference in how we treat the integration in this question compared to question 1 is that we must integrate over y first since the limits of y depend on x. Of course, we could work out how to exchange the limits by making the x limits depend on y, but there is no reason to do that here when the limits have been conveniently provided. We have ZZ Z 1 Z x V = e2x dA = dx dy e2x Z = 0 x2 0 R 1 x dx y e2x y=x2 = 2 Z 0 1 dx (x − x2 ) e2x . We will need to perform some integration by parts Z Z 1 1 1 1 1 1 1 1 2x 1 2x dx e2x = x e2x x=0 − e2x x=0 dx x e = x e x=0 − 2 2 2 4 0 2 0 2 e 1 1 e = −0− − = (e2 + 1) , 2 4 4 4 and Z 1 Z 1 2 2x 1 1 1 dx x e = x e x=0 − dx 2x e2x 2 2 0 0 e2 1 1 2 1 = −0− 2 (e + 1) = (e2 − 1) , 2 2 4 4 R1 where we made use of 0 dx x e2x again in the second integral. We now return to the volume, getting Z 1 1 1 1 V = dx (x − x2 ) e2x = (e2 + 1) − (e2 − 1) = . 4 4 2 0 2 2x 3. Prove that the integrals Z 2 Z x2 0 and 0 Z 4Z 0 (x2 − y) dy dx , 2 √ (x2 − y) dx dy , y have the same value. 3 Solution: The first integral can be evaluated as 2 Z 2 Z x2 Z 2 2 x y dx (x2 − y) dy dx = x2 y − 2 0 0 0 0 Z 2 Z 2 4 4 x x = x4 − dx dx = 2 0 0 2 2 32 16 x5 = = = . 10 0 10 5 The second integral can be evaluated as 2 Z 4 3 Z 4Z 2 x 2 dy − yx (x − y) dx dy = √ √ 3 y 0 0 y 3/2 Z 4 8 y = − 2y − − y 3/2 dy 3 3 0 4 Z 4 2y 3/2 8y 4y 5/2 8 2 − 2y + dy = −y + = 3 3 3 15 0 0 32 4(32) 160 − 240 + 128 48 16 = − 16 + = = = . 3 15 15 15 5 This values clearly agree. Note that the second integral is a bit harder. This shows how important it is to choose the order of integration carefully. 4. Use a double integral to find the volume of the solid in the first octant1 bounded by the surface 3x + y + 2z = 4 , which is a plane. Follow the steps outlined below: (a) Sketch the solid and its projection onto the xy-plane. 1 The first octant is the region x, y, z ≥ 0. 4 (b) Use the projection to find the limits of integration. (c) Perform the integration to find the volume. Solution: We will follow the suggested steps to work through the problem. (a) Note that we can rewrite the plane as 3 1 z = 2 − y − x. 2 2 The sketches of the solid and its projection onto the xy-axes are y 4 3 2 1 0 2.0 y 1.5 4 z 1.0 3 0.5 2 0.0 0.0 1 0.5 x 1.0 1.5 0.2 0.4 0.6 0.8 1.0 1.2 x Figure 2: The solid and its projection. (b) From the projection, we see that the lower limit of both variables is 0. The upper limit of one will depend on the other. Let’s choose to integrate one y first. Then the upper limit will come from the equation of the surface 3x + y + 2z = 4 onto the xy-plane, i.e. when z = 0. 5 Then we have 3x + y + 2(0) = 4 ⇒ y = 4 − 3x , which means the y limits are 0 and 4−3x. What about the limits in x? Well since the dependence on y has been dealt with, we can now integrate x over its entire range, which is 0 to 4/3. The 4/3 comes either from the picture of the projection or setting both y and z to 0 to get 3x + 0 + 2(0) = 4 ⇒ x = 4/3. We set y = 0 to find the upper limit of x since from the projection that is where x is largest. (c) We need to integrate over 2 − 12 y − 32 x. The volume is ZZ 1 3 V = 2 − y − x dA 2 2 R Z 4/3 Z 4−3x 3 1 = dx dy 2 − y − x 2 2 0 0 4−3x Z 4/3 1 1 = dx 2y − y 2 − xy 4 2 0 y=0 Z 4/3 1 1 = dx 2(4 − 3x) − (4 − 3x)2 − x(4 − 3x) − 0 4 2 0 Z 4/3 9x2 = dx 4 − 6x + 4 0 4/3 3x3 2 = 4x − 3x + 4 0 16 16 3 64 16 16 16 16 = −3 + −0= − + = . 3 9 4 27 3 3 9 9 6