# Surface integrals and surface areas of arbitrary parameter domains

I'm having trouble evaluating this surface integral. This would be very simple to solve if the parameter domain of the variables $u$ and $v$ was a square region.

However, that isn't the case here. I've tried using a change of variables and saying that $u = r \cos (x)$, and $v = r \sin (x)$. Where $0 < x < 2\pi$, and $0 < r < 2$ for the limits of integration.

However, computing the surface differential (absolute value of the cross product of the partial derivatives) using the new variables has become way to complicating and tedious. I end up with about $6$ or $7$ terms of sines and cosines under a square root that cannot be simplified.

Was this the way to approach this problem? I also thought about parametrizing the region R, but I'm not sure how that would work.

Also, since I'm already here, how would I be able to do this for some general region R in the $u-v$ plane? This is what lead me to think of a parametrization of a region R. So basically a parametrization within a parametrization. Mhmm, sounds interesting. Is that possible?

Thank you.

P.S. Tried looking around and haven't been able to find something that could answer my question.

EDIT: So I found the surface differential, but when I apply the the integral, am I suppose to add a jacobian since I am integrating in polar coordinates?

## 1 Answer

I think that part of the problem might be that you're trying too hard. First, you're using some definition to say that \begin{align} I &= \frac{1}{4} \int_S dS \\ &= \frac{1}{4} \int_u \int_v \| \frac{\partial \Phi}{ \partial u} \times \frac{\partial \Phi}{ \partial v} | ~du ~dv \end{align} where I've left the limits on $u$ and $v$ unwritten because they're a bit of a pain. Let's write out the integrand, though: \begin{align} \frac{\partial \Phi}{ \partial u} &=\begin{bmatrix} 1 \\ 2u \\ 1 \end{bmatrix} \\ \frac{\partial \Phi}{ \partial v} &=\begin{bmatrix} 1 \\ 2v \\ -1 \end{bmatrix} \\ \frac{\partial \Phi}{ \partial u} \times \frac{\partial \Phi}{ \partial v} &=\begin{bmatrix} -2u-2v \\ 2 \\ 2v-2u \end{bmatrix} \\ \end{align} The squared norm of that last item is \begin{align} \frac{\partial \Phi}{ \partial u} &=\begin{bmatrix} 1 \\ 2u \\ 1 \end{bmatrix} \\ \frac{\partial \Phi}{ \partial v} &=\begin{bmatrix} 1 \\ 2v \\ -1 \end{bmatrix} \\ \| \frac{\partial \Phi}{ \partial u} \times \frac{\partial \Phi}{ \partial v} \|^2 &= (2u+2v)^2 + (2u - 2v)^2 + 4\\ &= 4u^2 + 8uv + 4v^2 + 4u^2 - 8uv + 4v^2 + 4\\ &= 8u^2 +8v^2 + tu^2 + 4. \end{align} So that's your integrand, and the integral you want to find is

\begin{align} I &= \frac{1}{4} \int_S dS \\ &= \frac{1}{4} \int_u \int_v \sqrt{8u^2 +8v^2 + tu^2 + 4} ~du ~dv \end{align}

Now we change variables to $(r, x)$ as you suggested. The Jacobian determinant -- the determinant of the matrix of partials of $u$ and $v$ with respect to $r$ and $x$ is $$J = \left| \begin{matrix} \cos x & -r \sin x \\ \sin x & r \cos x\end{matrix}\right| = |r|$$ and the new integral is

\begin{align} I &= \frac{1}{4} \int_S dS \\ &= \frac{1}{4} \int_u \int_v \sqrt{8u^2 +8v^2 + tu^2 + 4} ~du ~dv \\ &= \frac{1}{4} \int_0^2 \int_0^{2\pi} \sqrt{8(r \cos x)^2 +8(r \sin x)^2 + 4} J ~dx ~dr \\ &= \frac{1}{4} \int_0^2 \int_0^{2\pi} \sqrt{8(r \cos x)^2 +8(r \sin x)^2 + 4} r ~dx ~dr \\ &= \frac{1}{4} \int_0^2 \int_0^{2\pi} \sqrt{8r^2 + 4} r ~dx ~dr\\ &= \frac{1}{4} \int_0^2 \int_0^{2\pi} 4\sqrt{2r^2 + 1} r ~dx ~dr\\ &= \int_0^2 \int_0^{2\pi} (2r^2 + 1)^{\frac{1}{2}} r ~dx ~dr \end{align} Substituting $s = 2r^2+1; ds = 4r dr; \frac{1}{4}ds = dr$ in that last integral, we get \begin{align} I &= \int_0^2 \int_0^{2\pi} (2r^2 + 1)^{\frac{1}{2}} r ~dx ~dr\\ &= \int_0^{2\pi} \left( \frac{1}{4}\int_{r =0}^2 {s}^{\frac{1}{2}} ~ds \right)~dx\\ &= \frac{2 \pi}{4}\int_{r =0}^2 {s}^{\frac{1}{2}} ~ds \\ &= \frac{\pi}{2}\int_{r =0}^2 {s}^{\frac{1}{2}} ~ds \\ &= \frac{\pi}{2}\left. \frac{2}{3} s^\frac{3}{2} \right|_{r =0}^2 \\ &= \frac{\pi}{2}\left. \frac{2}{3} (2r^2 + 1)^\frac{3}{2} \right|_{r =0}^2 \\ &= \frac{\pi}{2} \frac{2}{3} \left( (8 + 1)^\frac{3}{2} - 1)^\frac{3}{2} \right)\\ &= \frac{\pi}{2} \frac{2}{3} \left( 27 - 1 \right)\\ &= \frac{\pi}{2} \frac{2}{3} 26\\ &= \frac{26\pi}{3}. \end{align}

I probably made an algebra mistake or two in there, but that's the gist of the thing. Hope it helps.