Taking Multiple Surface Integrals

[TranscendArcu]

Homework Statement

http://img687.imageshack.us/img687/1158/skjermbilde20111204kl85.png

The Attempt at a Solution

I 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

r_y = <0,1,0>, r_θ = <-sinθ,0,cosθ>

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

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

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>

r_r = <cosθ,0,sinθ>, r_θ = <-rsinθ,0,rcosθ>

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

[tex]\int \int_{Q_2} <rcosθ,0,5> • <0,-r,0> d{Q_2} = 0[/tex]

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>

r_r = <cosθ,-cosθ,sinθ>, r_θ = <-rsinθ,rsinθ,rcosθ>

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

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

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

 

[mighty2000]

Thank you for sharing your solution! It looks like you have approached the problem correctly and have parametrized the different regions accurately. One thing to note is that in your second integral for Q2, the limits of integration should be from 0 to 1 instead of 0 to 2. Other than that, your solution looks good.

Another approach to this problem could be to use the divergence theorem and convert the surface integral into a volume integral. This may simplify the calculations and make the solution more efficient. However, your method is also valid and a good way to approach the problem.

Thanks again for sharing your solution and keep up the good work in your studies!A fellow scientist