I found this exercise while exercising for the exam:
Let $T$ $\subset$ $R^2$ be the triangle with these vertices $(0,0), (2,0), (0,1)$ and let $\Omega$ be the surface defined like this:
$\Omega$ = {$(x,y,z) \in R^3 : z^2 – x^2 – y^2 = 0, z > 0, (x,y) \in T$}
Evaluate $\iint_{\Omega}x^2 y dS $
I'm having a hard time solving it because it confounds me… I can't seem to "visualize" the situation. What's exactly the role of the triangular region, where am I going to have to use that region when solving the integral?
Could you help me visualize the problem in some way?
Best Answer
$T$ is not too hard to picture. It's a right triangle in the first quadrant of the $x,y$ plane with vertices at the given points.
$\Omega$ is the set of points in $\Bbb R^3$ on the given cone $z^2-x^2-y^2=0$ above the plane $z=0$, and further restricted to lie above $T$. To help visualize the cone, you can use level sets: take a slice of the cone at some plane $z=z_0$, then determine what kind of curve is left behind in the equation $z_0^2 - x^2 - y^2 = 0$. For example, at $z_0=\pm1$ we have $x^2+y^2=1$, the unit circle; at $z_0=\pm2$, $x^2+y^2=4$, a circle of radius $2$; and so on. In any plane $z=z_0$, we end up with a circle of radius $z_0$ centered on the $z$ axis. Joining them all together forms the surface shown in the figure below. The concentric circles you see along the walls of the cone are level curves.
The additional restriction of lying above $T$ means $\Omega$ should look like the figure below on the left, a cutaway of the cone in the shape of $T$ if you were to look directly down upon it. This is one of the roles of $T$: it tells us how to parameterize this cutaway. The figure on the right shows a bird's-eye view.
As for the integral, we can stay in Cartesian coordinates and use $z(x,y)=\sqrt{x^2+y^2}$ in our parameterization of the integration domain,
$$\Omega = \left\{\left(x,y,\sqrt{x^2+y^2}\right) \mid (x,y) \in T\right\}$$
This gives us two candidates for evaluating the surface integral,
$$\iint_\Omega x^2 y \, dS = \iint_T x^2 y \sqrt{\left(\frac{\partial z}{\partial x}\right)^2+\left(\frac{\partial z}{\partial y}\right)^2+1} \, dx\,dy = \begin{cases} \displaystyle \sqrt2 \int_0^2 \int_0^{1-\tfrac x2} x^2 y \, dy \, dx \\ \displaystyle \sqrt2 \int_0^1 \int_0^{2-2y} x^2 y \, dx \, dy \end{cases} = \boxed{\frac{2\sqrt2}{15}}$$
Just for the fun of considering a slightly more involved approach, we can convert to cylindrical coordinates. If $(x,y)=(r\cos\theta,r\sin\theta)$, then $z=\sqrt{x^2+y^2}=r$. The line forming the hypotenuse of $T$ transforms to
$$x+2y=2 \implies r(\cos\theta + 2\sin\theta) = 2 \implies r = \frac{2}{\sqrt5}\csc\left(\theta+\cot^{-1}2\right)$$
so that we can parameterize $\Omega$ by
$$\vec s(r,\theta) = \left\langle r\cos\theta,r\sin\theta, r\right\rangle \\ \Omega = \left\{\vec s(r,\theta) \mid 0\le r\le\frac2{\sqrt5}\csc\left(\theta+\cot^{-1}2\right) \land 0\le\theta\le\frac\pi2\right\}$$
The normal vector to $\Omega$ and the subsequent area element would be
$$\vec n = \frac{\partial \vec s}{\partial r} \times \frac{\partial \vec s}{\partial \theta} = \left\langle-r\cos\theta,-r\sin\theta,r\right\rangle \implies dS = \left\|\vec n\right\| dr\,d\theta = \sqrt2\,r\,dr\,d\theta$$
The subsequent integral agrees with our previous result,
$$\iint_\Omega x^2y \, dS = \sqrt2 \int_0^{\tfrac\pi2} \int_0^{\tfrac2{\sqrt5}\csc\left(\theta+\cot^{-1}2\right)} r^4 \cos^2\theta \sin\theta \, dr \, d\theta = \boxed{\frac{2\sqrt2}{15}}$$