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282 Midterm Solutions
- When I integrate
0
0
2 x− 2 y (x+y)^3 dxdy, I get 1. When I instead integrate^
0
0
2 x− 2 y (x+y)^3 dydx, I get −1. What’s going on here?
Fubini’s theorem would allow me to swap the bounds of integration, but it does not apply if the function in question is unbounded on the region of integration, as is the case here. Thus, I shouldn’t expect these two integrals to be equal.
- Compute:
(a)
0
x e y^2 dydx. Here, we need to swap the order of integration, as the inside integral is known to be impossible to compute. Drawing a picture, we see that D is a triangle with vertices (0,0), (0, 1) and (1,1). Thus, we can rewrite D = { 0 ≤ y ≤ 1 , 0 ≤ x ≤ y}, so the integral is now ∫ (^1)
0
∫ (^) y
0
ey
2 dxdy =
0
yey
2 dy
= e − 1 2 (b) the centroid of a lamina bounded by the parabola y = 4 − x^2 and the x-axis. Since the region is symmetric about the y-axis, it is clear that ¯x = 0. We need now only compute
m =
− 2
∫ (^4) −x 2
0
dydx
− 2
4 − x^2 dx
Mx =
− 2
∫ (^4) −x 2
0
y dydx
− 2
(4 − x^2 )^2 dx
− 2
16 − 8 x^2 + x^4 dx
y¯ = Mx m
(c)
D
x^2 + y^2 + z^2 dV where D is the region in the second octant (x < 0 , y > 0 , z > 0) bounded by the cones z =
x^2 + y^2 and z =
x^2 + y^2 and by the sphere of radius
This integral is most easily done in spherical coordinates, though it is not too bad in cylindrical. First, we need to write D in spherical. The octant gives θ between π 2 and π, while the sphere tells us that 0 ≤ ρ ≤ 3. The cones take a moment of thought, as we need to determine what angle φ determines each. The easiest way to do this (from my point of view) is to draw the slice of this shape lying in the y − z (or x − z) plane, which looks like a symmetric pair of ”wedges” with lines through the origin and a piece of a circle as a boundary. The lines are the cones and in this setting we can reduce the equations to, for example, z = y and z =
3 y. Now we just need the angle each of these lines makes with the z-axis, which we can solve using triangle trig. (Since, after all, tan(φ) = slope of these lines!). This gives π 6 ≤ φ ≤ π 4.
Now, we integrate, recognizing that the integrand is just ρ – another clue that spherical is good. ∫ π 4 π 6
∫ (^) π
π 2
0
ρρ^2 sin(φ) dρdθdφ =
∫ π 4 π 6 sin(φ) dφ
∫ (^) π
π 2 dθ
0
ρ^3 dρ
π 2
) (^81
- Set up, but do not compute, integrals for:
(a) the volume of a solid ball of radius 4 through the center of which has been drilled a cylindrical hole of radius 2. This integral is best done in cylindical coordinates, as the cylinder drilled out contains “straight lines”. We just set up the integral for the sphere, leaving out the r range cor- responding to the cylinderical hole. Since we’re computing volume, we use the integrand
V =
2
∫ (^2) π
0
∫ √ 4 −r 2
− √ 4 −r^2
r dzdθdr
(b) the integral of the function f (x, y, z) = z over the region bounded by the parabolas x = y^2 and y = x^2 and the planes z = 0 and z = x + y. This is just a matter of finding bounds. Drawing a picture of these two parabolic cylinders shows that they intersect at the points (0,0,z) and (1,1,z). If we choose x to be our constant bound, we can see from our picture that y is bounded below by x^2 and above by
x. z travels between the given bounds.
I =
0
∫ √x
x^2
∫ (^) x+y
0
z dzdydx
(c) the center of mass of a lamina in the shape of the section of an annulus of inner radius 1 and outer radius 2, between the lines y = x and y = −x and in the region x > 0, with density given by ρ(x, y) = xy + y^2. The annular section in question is just a polar rectangle, so that’s the way to go. After drawing a good picture – it should look like the region between the circle of radius 1 and that of radius 2 between the angles θ = − π 4 and θ = π 4 , we write down the integral.
Observe that the exponent of the e is −r^2 in polar, which will work out well with the r we get from the Jacobian. Further, this is an integral across the entire plane, so we can think of the polar region as θ ∈ [0, 2 π] and r from 0 to some distance R, which we will let go to infinity.
A^2 = lim R→∞
∫ R
0
∫ (^2) π
0
re−r 2 dθdr
= 2 π lim R→∞
∫ R
0
re−r 2 dr
= 2 π lim R→∞
∫ √R
0
e−u^ du
= π lim R→∞ (−e−
√R
= π A =
pi
