# Taking Multiple Surface Integrals

1. Dec 4, 2011

### TranscendArcu

1. The problem statement, all variables and given/known data
http://img687.imageshack.us/img687/1158/skjermbilde20111204kl85.png [Broken]

3. The attempt at a solutionI thought this was pretty hard and involved a number of different parts. Here's my work:

Let x=cosθ and z=sinθ, also let 0≤y≤2-x=2-cosθ. I parametrize Q1, which I define to be the cylindrical piece of the surface. Then, r(y,θ) = <cosθ,y,sinθ>, 0≤θ≤2*pi

ry = <0,1,0>, rθ = <-sinθ,0,cosθ>

<0,1,0> x <-sinθ,0,cosθ> = <cosθ,0,sinθ>

$$\int \int_{Q_1} <cosθ,y,5> • <cosθ,0,sinθ> d{Q_1}$$
$$\int_0 ^{2π} \int_0 ^{2-cosθ} cos^{2}θ + 5sinθ dydθ = π$$

Let x=rcosθ, z=rsinθ in the plane y=0. Also, let 0≤r≤1, 0≤θ≤2*pi. Let Q2 be the circle in the plane y=0. r(r,θ) = <rcosθ,0,rsinθ>

rr = <cosθ,0,sinθ>, rθ = <-rsinθ,0,rcosθ>

<cosθ,0,sinθ> x <-rsinθ,0,rcosθ> = <0,-r,0>

$$\int \int_{Q_2} <rcosθ,0,5> • <0,-r,0> d{Q_2} = 0$$

Let x=rcosθ, z=rsinθ, y=2-rcosθ. Also let 0≤r≤1, 0≤θ≤2*pi. Let Q3 be the region enclosed by the intersection of the cylinder with the x+y=2 plane. This surface is parametrized by r(r,θ) = <rcosθ,2-rcosθ,rsinθ>

rr = <cosθ,-cosθ,sinθ>, rθ = <-rsinθ,rsinθ,rcosθ>

<cosθ,-cosθ,sinθ> x <-rsinθ,rsinθ,rcosθ> = <-r,-r,0>

$$\int \int_{Q_3} <rcosθ,2-rcosθ,5> • <-r,-r,0> d{Q_3}$$
$$\int_0 ^{2π} \int_0 ^1 -r^2cosθ - 2r +r^2cosθ drdθ$$
$$\int_0 ^{2π} (-1/3)r^{3} * cosθ -r^2 + (1/3)r^{3}*cosθ| ^{r=1} _{r=0} dθ = -2π$$

But I change the sign due to orientation. So 2*pi. Adding all the piece together, I have 3*pi.

Last edited by a moderator: May 5, 2017