Solutions to Homework from Sections 12.4, 12.6
Math 234, Spring 2008
Section 12.4
10.
ZZR
(x+y)dA =Z3π/2
π/2Z2
1
(rcos θ+rsin θ)r dr dθ = Z3π/2
π/2
(cos θ+ sin θ)dθ!Z2
1
r2dr
=sin θ−cos θ|3π/2
π/2 r3
3
2
1!= (−1−0−1 + 0) 8
3−1
3=−14
3
20. The paraboloid z= 1 + 2x2+ 2y2meets the plane z= 7 where 7 = 1 +2x2+ 2y2=⇒2x2+ 2y2= 6 =⇒x2+y2= 3,
so the domain of integration is the part of the circle about the origin of radius √3 in the first quadrant.
Volume = Zπ/2
0Z√3
0
(7 −(1 + 2r2)) r dr dθ =π
2Z√3
0
(6r−2r3)dr =π
2 3r2−1
2r4
√3
0!=π
29−9
2=9π
4
30. (a)
Z2π
0ZR
0
e−rr dr dθ = 2πZR
0
e−rr dr
We can do this integral using integration by parts, or we can look it up in the table of integrals:
Volume = 2πZR
0
e−rr dr = 2π(−r−1)e−r
R
0= 2π((−R−1)e−R+ 1) = 2π(1 −e−R−Re−R) ft3
(b) The average volume per square foot is obtained by dividing the total volume by the area of the circle of radius R:
Average = 2π(1 −e−R−Re−R)
πR2=2(1 −e−R−Re−R)
R2ft3/ft2
Section 12.6
10. We can view this as the graph of x=f(y, z) = y2+z2, so the surface area is given by:
ZZRs1 + ∂f
∂y 2
+∂f
∂z 2
dy dz =ZZRp1+4y2+ 4z2dy dz
To do this integral over the disk y2+z2≤9, we convert to polar coordinates:
ZZRp1+4y2+ 4z2dy dz =Z2π
0Z3
0p1+4r2r dr dθ = 2πZ3
0p1+4r2r dr = 2π1
8 2
3(1 + 4r2)3/2
3
0!=π
6(373/2−1)
20. (a) ~r(u, v) = hau cos v, bu sin v , u2i. So ~ru=hacos v, b sin v, 2uiand ~rv=h−au sin v , bu cos v, 0i. This means that
~ru×~rv=h−2bu2cos v, −2au2sin v, abu cos2v+abu sin2vi=h−2bu2cos v , −2au2sin v, abui. So |~ru×~rv|=
p4b2u4cos2v+ 4a2u4sin2v+a2b2u2=up4b2u2cos2v+ 4a2u2sin2v+a2b2. This means that the surface area
is given by:
Z2π
0Z2
0
up4b2u2cos2v+ 4a2u2sin2v+a2b2du dv
(b) Since x
a=ucos vand y
b=usin v, we observe that:
x
a2+y
b2=u2cos2v+u2sin2v=u2=z
