How to change variables in surface integral

Question

Now I know how to change variable of multiple integral, and I'd like to find out how to do this on surface integral. Like this $$\iint_\Sigma x^2 \,dy\,dz+y^2\,dz\,dx+z^2\,dx\,dy,\,where \Sigma:(x-1)^2+(y-1)^2+\frac{z^2}{4}=1(y>1) $$ I think it would be convenient if let $$u=x-1,v=y-1,w=\frac{z}{2} $$ It's just an example, you don't have to solve this.

Answer 1

$\displaystyle \iint_S x^2 \,dy\,dz+y^2\,dz\,dx+z^2\,dx\,dy,\ $
where $S:(x-1)^2+(y-1)^2+\frac{z^2}{4}=1(y>1)$

Please note,

$\displaystyle \iint_S x^2 \,dy\,dz+y^2\,dz\,dx+z^2\,dx\,dy = \iint_S \vec{F} \cdot \hat{n} \ dS , $ where $\vec{F} = (x^2,y^2,z^2)$.

If you parametrize the surface in spherical coordinates as,

$r(\theta, \phi) = (1 + \cos\theta \sin\phi, y = 1 + \sin\theta \sin \phi, z = 2\cos\phi)$, the normal vector is $(2\cos\theta \sin^2\phi, 2\sin\theta \sin^2\phi, \sin\phi \cos\phi); 0 \leq \phi \leq \pi, 0 \leq \theta \leq \pi$ (Note $y \geq 1)$

$\vec{F} = ((1+\cos\theta\sin\phi)^2,(1+\sin\theta\sin\phi)^2,4\cos^2\phi)$

Now you do the dot product and integrate.

Or you could parametrize in cylindrical coordinates as $(1+\rho\cos\theta, 1+\rho\sin\theta, 2 \sqrt{1-\rho^2})$.

Or in cartesian, $u = x-1, v = y-1, w = \frac{z}{2}$.

But it may be a bit easier using divergence theorem,

To apply divergence theorem, we close the surface by adding an elliptic disc at $y=1$. We can subtract flux through it later.

$\nabla \cdot \vec{F} = 2(x+y+z)$

Using $u = x-1, v = y-1, w = \frac{z}{2}, |J| = 2$, the region has transformed to $R: u^2+v^2+w^2 = 1, v \geq 0$ which is a hemisphere of unit radius.

Divergence is rewritten as $2(x+y+z) = 2(2+u+v+2w)$

So the volume integral to find flux through closed surface is,

$\displaystyle \iiint_R 2(2+u+v+2w) \ |J| \ du \ dv \ dw$

Now we note that $u$ and $w$ are both odd function and due to symmetry wrt $VW$ plane and $UV$ plane respectively, the integral will be zero. The reason the integral is not zero for $v$ as our transformed surface is hemisphere with $v \geq 0$. So it boils down to,

$8 \displaystyle \iiint_R \ du \ dv \ dw + 4 \displaystyle \iiint_R v \ du \ dv \ dw$

First part is simply $8$ times volume of the hemisphere of radius $1$. Now you can find the second part and add.

Then subtract flux through elliptic disc that we added. Normal vector for the disc is $(0,-1,0)$ and $y = 1$.

$\vec{F}\cdot\hat{n} = (x^2,y^2,z^2) \cdot (0,-1,0) = - 1 \ (y = 1)$.

Ellipse has semi axes $a = 1, b = 2$ and its area is $2\pi \ $ (using $\pi ab)$. The flux through the surface is $-2\pi$.