Faster way to calculate the area of this surface

I have to solve this exercise:

Find the area of the portion of the surface $z=xy$ included between the two cylinders $x^2+y^2=1$ and $x^2+y^2=4$

what i did so far:

I parameterized the surface using cylindircal polar coordinates $$\ \Phi(\rho,\theta)=\ \begin{pmatrix}\ \rho\cos(\theta) \\ \rho\sin(\theta) \\ \rho^2\cos(\theta)\sin(\theta) \end{pmatrix},\ \ \ \rho\in[1,2], \ \ \theta\in[0,2\pi]$$

then i computed the partial derivates of $\Phi$

$$\Phi_\rho=\ \begin{pmatrix}\ \cos(\theta) \\ \sin(\theta) \\ 2\rho\cos(\theta)\sin(\theta) \end{pmatrix}\ \text{ and }\ \Phi_\theta=\ \begin{pmatrix}\ -\rho\sin(\theta) \\ \rho\cos(\theta) \\ \rho^2(\cos^2(\theta)-\sin^2(\theta)) \end{pmatrix}\$$

At this point i knew the answer should be $$\int_0^{2\pi}d\theta\int_1^2|\Phi_\rho\times\Phi_\theta|d\rho$$

the problem is that the expression $|\Phi_\rho\times\Phi_\theta|$ is massive and it would take me forever to compute its integral. Since this question is supposed to be answered within 3 minutes there must be some kind of trick i can use to do it.

EDIT:

The correct answer is $2\pi(5\sqrt{5}-2\sqrt{2})/3$ but i don't know how to get there

• Within three minutes? How?! I am going to follow this thread, I'm interested in what will people come up with – KKZiomek May 21 '18 at 13:23
• @KKZiomek yes, it comes from a recent exam of my multivariable calculus course in which there are 9 questions and the time limit is 30 minutes – Censacrof May 21 '18 at 13:28
• Is it possible that we talk about signed area and this cancels due to symmetry? – mvw May 21 '18 at 13:37
• @mvw no, i know that the answer is $2\pi(5\sqrt{5}-2\sqrt{2})/3$ but i don't know how to get there – Censacrof May 21 '18 at 13:41

It is actually pretty simple. The trick is compute the area element before changing variable to cylindrical polar coordinate $(x,y,z) = (\rho\cos\theta,\rho\sin\theta,z)$.

\begin{align}\verb/Area/ &= \int_{1 \le \sqrt{x^2+y^2} \le 2}\sqrt{1+\left(\frac{\partial z}{\partial x}\right)^2 + \left(\frac{\partial z}{\partial y}\right)^2} dxdy\\ &= \int_{1 \le \sqrt{x^2+y^2} \le 2}\sqrt{1+y^2 + x^2} dxdy\\ &= \int_0^{2\pi} \int_1^2\sqrt{1+\rho^2} \rho d\rho d\theta\\ &= \frac{2\pi}{3} \left[ \sqrt{1+\rho^2}^3\right]_{1}^2\\ &= \frac{2\pi}{3}\left(5\sqrt{5}-2\sqrt{2}\right) \end{align}

• first of all thanks for the answer, i have a few questions: what is $z_x$ and $z_y$? Why the integrand function is $\sqrt{1+z_x^2+z_y^2}$? Sorry if its supposed to be obvious but i'm pretty new to multivariable integrals. – Censacrof May 21 '18 at 14:05
• @Censacrof $z_x, z_y$ are shorthand for $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ which is relatively common for this sort of application. – achille hui May 21 '18 at 14:07
• thanks again, but still i don't get the resoning behind the $\sqrt{1+z_x^2+z_y^2}$ bit, i'd be very gratefull to you if you provided the reasoning you made to come to your conclusion – Censacrof May 21 '18 at 14:12
• @Censacrof Let's say $\Phi(x,y) = (x,y,z(x,y))$, then $\Phi_x = (1,0,z_x)$, $\Phi_y = (0,1,z_y)$ and $\Phi_x \times \Phi_y = ( -z_x, -z_y, 1) \implies |\Phi_x \times \Phi_y| = \sqrt{1 + z_x^2 + z_y^2}$. – achille hui May 21 '18 at 14:18
• oh i get it now, thank you very much for your patience – Censacrof May 21 '18 at 14:20

It's actually not that hard to work out your cross product, using $Cos(\theta)^2 + Sin(\theta)^2 = 1$:

$|\Phi_\rho \times \Phi_\theta| = \begin{vmatrix}\rho^2*(Cos(\theta)^2-Sin(\theta)^2)Sin(\theta)-2\rho^2*Cos(\theta)^2Sin(\theta)\\-2\rho*Cos(\theta)Sin(\theta)^2 -\rho^2*(Cos(\theta)^2-Sin(\theta)^2)Cos(\theta)\\ \rho*Cos(\theta)^2 + \rho*Sin(\theta)^2\end{vmatrix} = \begin{vmatrix}-\rho^2*Sin(\theta)^2\\-\rho^2*Cos(\theta)^2\\\rho\end{vmatrix} = \sqrt{\rho^4 + \rho^2} = \rho*(\rho^2+1)^{1/2}$. The integration gives $2\pi[\frac{1}{3}*(\rho^2+1)^{3/2}]^2_1 = \frac{2\pi}{3}(5^{3/2}-2^{3/2})$.

• Pretty sure those first two components should be proportional to $\sin \theta$ and $\cos \theta$ respectively, not $\sin^2 \theta$ and $\cos^2\theta$. Otherwise, nice work. – Michael Seifert May 21 '18 at 18:11