I'm working on a problem that states:

Evaluate $\iint_S (x^2z+y^2z)\,dS$ where $S$ is part of the plane $z=4+x+y$ that lies inside the cylinder $x^2+y^2=4$.

I want to use the following surface integral formula:

$$\iint_S f(x,y,z)\,dS = \iint_D f\left(\vec r\left(u, v\right)\right) \left\lVert\vec r_u \times \vec r_v \right\rVert \,dA$$

It requires the surface to be parametrized, however. How can I parameterize the surface that was given in this problem to start?

EDIT: I was thinking of going polar.

$$ r^2 = 4 $$ $$ r = 2 $$ $$ z = 4 + 2cos(\theta)+2sin(\theta) $$


The surface $S$ is the part of the plane $z=4+x+y$ that lies inside the cylinder $x^2+y^2=4$. We want to describe $S$ as a parametric surface $\vec r=\vec r(u,v)$. This is most easily accomplished in cylindrical coordinates, Where the equation of the plane takes the form $z=4+\rho\cos{\phi}+\rho\sin{\phi}=4+\rho(\cos{\phi}+\sin{\phi})$, and the equation of the cylinder takes the form $\rho=2$.

We have the following parametric representation of $S$:


Computing the partial derivatives and their cross product, we get




Thus, $\|\vec{r}_\rho\times\vec{r}_\phi\|=\sqrt{3}\rho$.

Rewriting the integrand in cylindrical coordinates, we have


Recall that on the surface $S$, we have $z=4+\rho(\cos{\phi}+\sin{\phi})$, so then


Putting it all together, the surface integral evaluates to

$$I=\iint_{S}f(\vec{r})\,dS=\iint_{D}f(\vec{r}(\rho,\phi))\|\vec{r}_\rho\times\vec{r}_\phi\|\,dA\\ =\sqrt{3}\int_{0}^{2\pi}\int_{0}^{2}\rho^3(4+\rho(\cos{\phi}+\sin{\phi}))\,d\rho\,d\phi$$

We can make the evaluation of the above integral even easier by switching the order of integration and noting that $\cos\phi+\sin\phi$ is periodic in $\phi$ with period $2\pi$ and that the integral over one period is automatically zero. Hence, the value of the surface integral is:

$$I=\sqrt{3}\int_{0}^{2\pi}\int_{0}^{2}\rho^3(4+\rho(\cos{\phi}+\sin{\phi}))\,d\rho\,d\phi=\sqrt{3}\int_{0}^{2}\int_{0}^{2\pi}(4\rho^3+\rho^4(\cos{\phi}+\sin{\phi}))\,d\phi\,d\rho\\ =\sqrt{3}\int_{0}^{2}\int_{0}^{2\pi}4\rho^3\,d\phi\,d\rho\\ =2\sqrt{3}\pi\int_{0}^{2}4\rho^3\,d\rho\\ =2\sqrt{3}\pi(2^4)\\ =2^5\sqrt{3}\pi.$$


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