surface area of f(x,y) on a circular region

Question

Find Surface area of $f(x,y) = xy$ over the region $\{(x,y) : x^2 + y^2 \le 16\}$

I know that I will probably need to convert to polar

I have some notes that show that if the area was a rectangle then I would do a double integral over that region and the function in the middle would be $\sqrt{ 1 +(\partial_x)^2 + (\partial_y)^2}$

If someone can show an example of how to do a similar problem with different numbers, that would be amazing

Answer 1

The surface area is given by $$S_A=\iint_R \sqrt{1+(f_x)^2+(f_y)^2}dA.$$ In this case $R=\{ (x,y) \in \mathbb{R}^2:x^2+y^2 \leq 16 \}$, $f_x(x,y)=y$ and $f_y(x,y)=x$. Then $$S_A=\iint_{R} \sqrt{1+x^2+y^2}dA.$$ Using polar coordinates we get that $r \in [0,4]$ and $\theta \in [0,2\pi)$, and so $$S_A=\int_0^{2\pi}\int_0^4 r\sqrt{1+r^2}drd\theta.$$ It's easy to see that $$\int_0^x r\sqrt{1+r^2}dr=\frac{(x^2+1)^{3/2}}{3} \implies \int_0^4 r\sqrt{1+r^2}dr=\frac{17\sqrt{17}-1}{3}$$ and so $$S_A=\int_0^{2\pi} \frac{17\sqrt{17}-1}{3}d\theta=\frac{2\pi}{3}\left(17\sqrt{17}-1 \right).$$

Answer 2

Note the magnitude of your surface normal is given as $$ \sqrt{(f_x)^2+(f_y)^2+1}=\sqrt{y^2+x^2+1} $$ and thus your surface integral is $$ \int\int_{D}\sqrt{y^2+x^2+1}\mathrm dx \mathrm dy $$ where $D$ is the disc of radius $4$ around the origin, all of which is all very convenient for polar, where your integral becomes $$ \int_0^{2\pi}\int_0^4r\sqrt{1+r^2}\mathrm dr \mathrm d\theta=2\pi\int_0^4r\sqrt{1+r^2}\mathrm dr\\ =\pi\int_1^{17}\sqrt{u}\mathrm du $$