# Double integral surface area within a cylinder

Example: Find the area of the part of the surface $z-xy=\pi$ that lies within the cylinder $x^2+y^2=16$.

I'm having trouble visualising this problem. I was thinking of using the double integral equation for surface area on the equation $z-xy=\pi$ but I wasn't sure how to set up the bounds of the double integral and am not sure whether or not to convert to polar coordinates. How should I get started?

Thanks!

Yes, it seems this should be converted to polar coordinates. We have a circular area, so it makes sense. The circle is at the origin with a radius of 4, so the bounds for $r$ are from $0$ to $4$. Then, because it's a complete circle, we integrate $\theta$ from $0$ to $2\pi$. Then, we have $z-xy=\pi$ that needs to be derived before being converted to polar coordinates:
$z=\pi+xy$
$z_x = y = r\sin(\theta), x_y = x = r\cos(\theta)$

Final integral: $\int_0^{2\pi} \int_0^4 \sqrt{1+(z_y)^2+(z_x)^2} = \int_0^{2\pi} \int_0^4 \sqrt{1+r^2\sin^2(\theta)+r^2\cos^2(\theta)}$
$=\int_0^{2\pi} \int_0^4 \sqrt{1+r^2} \ dr d\theta$

Now here's some really helpful advice. This integral is about to turn ugly. You will have to do a trig substitution: Let $r = \tan(\alpha)$. Then $dr = \sec^2(\alpha)$ and we have $$\int_0^{2\pi} d\theta \cdot \int_{r=0}^{r=4} \sec^3(\alpha) \ d\alpha$$.
Have fun, and good luck!

• Could you also use this formula for surface area? $\int \int \sqrt{1+f_x^2+f_y^2}$. Do you also have to add the integration constant, $r$, for polar coordinates when putting together this double integral? – Gerardo Suarez May 1 '18 at 2:28
• ah shoot! it's surface area! – Christopher Marley May 1 '18 at 2:35
• I've redone the answer. Can't believe i solved for volume first! – Christopher Marley May 1 '18 at 2:40