18.022 Recitation Handout (with solutions)
13 November 2014
x/2+1
Z 2Z
2
x5 (2y − x)e(2y−x) dy dx by making the substitu-
1. (5.5.10 in Colley) Evaluate the integral
x/2
0
tion u = x and v = 2y − x.
Solution. Substitution shows that the limits of integration become u ∈ [0, 2] and v ∈ [0, 2]. We
calculate
∂(x, y) 1 1/2 ∂(u, v) = 0 1/2 = 1/2,
so the formula for change of variables gives
Z 2Z 2
Z 2 1 1 5 v2 2
5 v2 ∂(x, y) u ve dv
du
=
du
u
e
∂(u, v) 0
0
0
0 2 2
Z 2
1 5 4
8
=
u (e − 1) du = (e4 − 1) .
4
3
0
2. Let D be a parallelogram with vertices (0, 0), (1, 0), (1, 1), and (2, 1). Calculate
ways:
(a) Find
(b) Find
!
D
!
D
!
D
1 dA in two
1 dA without using calculus.
1 dA using the change of variables u = 2x − 2y and v = 2y.
Solution. (a) The integral in question is the area of the parallelogram, which is (base)(height) =
1 × 1 = 1 . (b) To calculate this area using the suggested change of variables, we note that the
linear transformation sends the four vertices of D to the vertices of a square [0, 2] × [0, 2]. Since the
transformation is linear, it sends D to [0, 2] × [0, 2]. We calculate
"
Z 2 Z 2 ∂x/∂u ∂y/∂u du dv
1 dA =
1
∂x/∂v ∂y/∂v D
0
0
Z 2 Z 2 1
2
0 =
1 1
du dv
1/2 0
0
2
= 4(1/2) = 1 .
2
3. (5.5.30 in Colley) Find the volume of the solid that is bounded by the paraboloid z = 9 − x2 − y2 ,
the xy-plane, and the cylinder x2 + y2 = 4.
Solution. To find the volume of the region, we integrate 1 over the region. We use cylindrical
coordinates:
Z 2π Z 2 Z 9−r2
Z 2
volume =
1 r dz dr dθ = (2π)
(9 − r2 )r dr. = 28π .
0
0
0
0
4. (5.5.29 in Colley) Find the volume of the region W that represents the intersection of the solid
cylinder x2 + y2 ≤ 1 and the solid ellipsoid 2(x2 + y2 ) + z2 ≤ 10.
Solution. We integrate in cylindrical coordinates:
√
1 Z 2π Z
Z
0
0
10−2r2
√
− 10−2r2
Z
r dz dr dθ = 2π
0
1
√
4√ √
2r 10 − 2r2 dr =
2 5 5−8 π .
3