MA 242 Test 3 Solutions1. [10 points] If f is a constant function, f(x, y) = k, and R = [a, b] × [c, d], computeZZRkdA.Which theorem was used in this problem?Solution:ZdcZbakdxdy = kZdcdyZbadx = k · [y]dc· [x]ba= k(d − c)(b − a)We had to use Fubini’s Theorem to compute this integral.2. [15 points] Evaluate the integral,Z10Z1√ypx3+ 1dxdy, by reversing the order of integration.Solution:If we look at the integral in its given Type II form we see thatD = {(x, y)|0 ≤ y ≤ 1,√y ≤ x ≤ 1}.This region can also be described in Type I form. In Type I form,D = {(x, y)|0 ≤ x ≤ 1, 0 ≤ y ≤ x2}.ThenZ10Z1√ypx3+ 1dxdy =Z10Zx20px3+ 1dydxFirst compute the inner integral,Rx20√x3+ 1dy = [y√x3+ 1]y=x2y=0= x2√x3+ 1Next compute the outer integral via u-substitution,(choose u = x3+ 1, then du = 3x2dx)Z10x2px3+ 1dx = [29(x3+ 1)3/2]x=1x=0=29(23/2− 1)3. [15 points] Find the moment of inertia about the x-axis for the lamina that occupies the region,D = {(x, y)|0 ≤ y ≤ ex, 0 ≤ x ≤ 1} and has the density function ρ(x, y) = y.Solution:The moment of inertia about the x-axis is Ix=ZZDy2ρ(x, y)dA =Z10Zex0y3dydxCompute the inner integral,Zex0y3dy = [y44]ex0=14e4xNext compute the outer integral,14Z10e4xdx =116[e4x]10=116[e4− 1]4. [10 points] Set up the triple integral in rectangular coordinates that is needed to find the volume ofthe solid enclosed by the cylinder x2+ y2= 9 and the planes y + z = 5 and z = 1. Set up E as a typeI solid region and set up D as a type I plane region. Do not compute this integral.Solution:First note that V (E) =ZZZEdVThe solid region E is of type I and D ,its projection onto the xy-plane, is a type I region(D is circle of radius 3 centered at the origin).So E = {(x, y, z)|5 − y ≤ z ≤ 1, −√9 − x2≤ y ≤√9 − x2, −3 ≤ x ≤ 3}Then V =Z3−3Z√9−x2−√9−x2Z5−y1dzdydx5. [10 points] Set up the triple integral in cylindrical coordinates that would be used to evaluateZZZEx2dV ,where E is the solid that lies within the cylinder x2+ y2= 1, above the plane z = 0, and below thecone z2= 4x2+ 4y2. Do not compute this integral.Solution:In cylindrical coordinates, E is bounded by the cylinder r = 1, the plane z = 0, and thecone z = 2r.So E = {(r, θ, z)|0 ≤ θ ≤ 2π, 0 ≤ r ≤ 1, 0 ≤ z ≤ 2r}SoZZZEx2dV =Z2π0Z10Z2r0r2cos2θ rdzdrdθ6. [10 points] Find the Jacobian of the transformation x = α sin β , y = α cos β.Solution:∂(x, y)∂(α, β)=∂x/∂α ∂x/∂β∂y/∂α ∂y/∂β=sin β α cos βcos β −α sin β= −α sin2β − α cos2β = −α7. [10 points] On the back of the test.Solution:a) Type 1b) Type 2c) Type 1d) Type