Now because our surface is a cylinder we’ll need to parameterize it and use the following formula for the surface integral.

$$\iint\limits_{S}{{f\left( {x,y,z} \right)\,dS}} = \iint\limits_{D}{{f\left( {\vec r\left( {u,v} \right)} \right)\left\| {{{\vec r}_u} \times {{\vec r}_v}} \right\|\,dA}}$$

where $u$ and $v$ will be chosen as needed when doing the parameterization.

We saw how to parameterize a cylinder in the previous section so we won’t go into detail for the parameterization. The parameterization is,

$$\vec r\left( {x,\theta } \right) = \left\langle {x,2\sin \theta ,2\cos \theta } \right\rangle \hspace{0.25in}\,\,\,\,\,0 \le \theta \le 2\pi ,\,\,\,\,\,0 \le x \le 3 - z = 3 - 2\cos \theta$$

We’ll use the full range of $\theta$ since we are allowing it to rotate all the way around the x‑axis. The $x$ limits come from the two planes that “bound” the cylinder and we’ll need to convert the upper limit using the parameterization.

Next, we’ll need to compute the cross product.

$${\vec r_x} = \left\langle {1,0,0} \right\rangle \hspace{0.25in}{\vec r_\theta } = \left\langle {0,2\cos \theta , - 2\sin \theta } \right\rangle$$ $${\vec r_x} \times {\vec r_\theta } = \left| {\begin{array}{*{20}{c}}{\vec i}&{\vec j}&{\vec k}\\1&0&0\\0&{2\cos \theta }&{ - 2\sin \theta }\end{array}} \right| = 2\sin \theta \vec j + 2\cos \theta \vec k$$

The magnitude of the cross product is,

$$\left\| {{{\vec r}_x} \times {{\vec r}_\theta }} \right\| = \sqrt {4{{\sin }^2}\theta + 4{{\cos }^2}\theta } = 2$$

The integral is then,

$$\iint\limits_{S}{{2y\,dS}} = \iint\limits_{D}{{2\left( {2\sin \theta } \right)\left( 2 \right)\,dA}} = \iint\limits_{D}{{8\sin \theta \,dA}}$$

Don’t forget to plug the $y$ component of the surface parametrization into the integrand and $D$ is just the limits on $x$ and $\theta$ we noted above in the parameterization.

Show Step 3

Now all that we need to do is evaluate the double integral and that shouldn’t be too difficult at this point.

The integral is then,

$$\begin{align*}\iint\limits_{S}{{2y\,dS}} & = \int_{0}^{{2\pi }}{{\int_{0}^{{3 - 2\cos \theta }}{{8\sin \theta \,dx}}\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{\left. {\left( {8x\sin \theta } \right)} \right|_0^{3 - 2\cos \theta }\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{8\left( {3 - 2\cos \theta } \right)\sin \theta \,d\theta }}\\ & = \int_{0}^{{2\pi }}{{24\sin \theta - 16\sin \theta \cos \theta \,d\theta }}\\ & = \int_{0}^{{2\pi }}{{24\sin \theta - 8\sin \left( {2\theta } \right)\,d\theta }}\\ & = \left. {\left( { - 24\cos \theta + 4\cos \left( {2\theta } \right)\,} \right)} \right|_0^{2\pi } = \require{bbox} \bbox[2pt,border:1px solid black]{0}\end{align*}$$