Answer Key for Quiz 7 (section B)
1. The top half of the loop of y2=x2+x5looks like
It intersects the x-axis at (−1,0) and at (0,0). A vertical strip of this region at a point xgenerates a
circular disk of radius √x2+x5when revolved around the x-axis. Since the area of such a disk is π(x2+x5),
the volume we get is obtained by multiplying by dx and integrating over the possible values of x, namely
−1≤x≤0. This gives
V=Z0
−1
π¡x2+x5¢dx =πµx3
3+x6
6¶¯
¯
¯
¯
0
−1
=π·(0 + 0) −µ−1
3+1
6¶¸=πµ1
3−1
6¶=π
6.
As with the volume of the week, horizontal strips are hopeless with this curve, because you can’t solve for
xin terms of y.
2. The curves y= cos xand y= sin xintersect at (π/4,1/√2), so the region of interest looks like
A vertical strip at a given xgenerates a washer with outer radius cosxand inner radius sin xwhen revolved
around the x-axis. The area of such a washer is
π(cos x)2−π(sin x)2=π¡cos2x−sin2x¢.
Multiplying by dx and integrating over all the possible values of xwe get
V=πZπ
4
0¡cos2x−sin2x¢dx.
If you happen to know that cos2x−sin2x= cos 2x, that makes this easier:
V=πZπ
4
0
cos 2x dx =π
2sin 2x¯
¯
¯
π
4
0
=π
2³sin π
2−sin 0´=π
2(1 −0) = π
2.
Otherwise you can use #17 and #18 in the table with n= 2 to get
V=π·x
2+sin xcos x
2−µx
2−sin xcos x
2¶¸¯
¯
¯
¯
π
4
0
=π[sin xcos x]|
π
4
0=πµ1
√2
1
√2−0·1¶=π
2.
Doing this problem with cylindrical shells is possible, but quite difficult.
