Sample solutions to Sec 16.4-16.5 worksheet.
1. Answer
Z2π
0Z4
0
sin(r2)rdrdθ =π(cos(0) −cos(16)) = π(1 −cos(16)).
2. This is way too difficult because of the expression that I picked.
Sorry.
3 Intersection is a circle of radius 4 over the xy plane so use polar
coordinates Z2π
0Z2
0
(4 −r2)rdrdθ
4. This is similar to the problem I did in class during the review on
Tue.
5. The volume is bounded above by z= 20 −x2−y2and bounded
below by z=x2+y2. Now the two intersect in x2+y2= 10, so we
have a disc of radius √10. Therefore in polar coordinates
V olume =Z2π
0Z√10
0
[(20 −r2)−(r2)]rdrdθ
this sets up the integral. The rest is just computations.
6. Probably the easiest thing to do is switch to polat coordinates
there the integral is
Zπ/4
0Z2/cos θ
1/cos θ
1
rrdrdθ =Zπ/4
0
1
cos θ= ln(sec(θ) + tan(x)) |π/4
0
= ln(1/√2 + 1) −ln(1 + 0) = ln(1/√2 + 1).
7. This is similar to number 5, I think it will be easiest if you move
everything up by 1 (right now z=px2+y2−1 is a cone starting at
(0,0,−1).) then use spherical coordinates (this will be just like the ice
cream cone).
8. The two surfaces intersect at x2+y2= 9/2 so it is a disc of radius
3/√2. If I use cylindrical coordinates I have
V olume =Z2π
0Z3/√2
0Z√9−r2
r
1rdzdrdθ.
The rest is just computations.
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