Note. In each example, the illustration shows the surface. The part of the surface highlighted in orange is the part whose area you are asked to find.
Find the area of the surface $\definecolor{answer}{RGB}{0,0,230}\mathbf{r}(u,v) = \langle u\cos v,u\sin v,v \rangle$
$0\le u\le 3$, $-\pi \le v \le \pi$.
Find the area of the part of the plane $x+2y+z=4$
that lies in the first octant.
Find the surface area of the cylinder $x^2+y^2=4$
between $z=-1$ and $z=3$.
Find the area of the part of the surface
$z=\dfrac{xy}{2}$
contained in the cylinder $x^2+y^2=4$.

### Solution

\begin{align*} \mathbf{r}_u &= \langle \cos v, \sin v, 0 \rangle \\ \mathbf{r}_v &= \langle -u\sin v, u\cos v, 1 \rangle \\ \left|\mathbf{r}_u\times\mathbf{r}_v\right| &= \left|\langle \sin v, -\cos v, u\rangle\right| = \sqrt{1+u^2} \\ \\ A(S) &= \int_{-\pi}^\pi \int_0^3 \sqrt{1+u^2}\,du\,dv = \int_{-\pi}^\pi\,dv\int_0^3\sqrt{1+u^2}\,du \\ &= 2\pi\bigg[ \frac{1}{2}u\sqrt{1+u^2}+\frac{1}{2}\ln\left|u+\sqrt{1+u^2}\right| \bigg]_0^3 = \color{answer}\mathbf{\pi(3\sqrt{10}+\ln(3+\sqrt{10}))} \end{align*} \begin{align*} z &= 4-x-2y \\ \\ A(S) &= \int_0^2\int_0^{4-2y}\sqrt{1+\left( \frac{\partial z}{\partial x} \right)^2 + \left( \frac{\partial z}{\partial y} \right)^2}\,dx\,dy = \int_0^2\int_0^{4-2y}\sqrt{1+(-1)^2+(-2)^2}\,dx\,dy \\ &= \int_0^2\int_0^{4-2y}\sqrt{6}\,dx\,dy = \sqrt{6}\int_0^2(4-2y)\,dy \\ &= \sqrt{6}\bigg[4y-y^2\bigg]_0^2 = \color{answer}\mathbf{4\sqrt{6}} \end{align*} Use cylindrical coordinates to parametrize the surface. \begin{align*} \mathbf{r}(\theta,z) &= \langle 2\cos\theta,2\sin\theta,z\rangle,\quad 0\le \theta\le 2\pi,~-1\le z\le 3 \\ \mathbf{r}_\theta &= \langle -2\sin\theta,2\cos\theta,0\rangle \\ \mathbf{r}_z &= \langle 0,0,1 \rangle \\ \left|\mathbf{r}_\theta\times\mathbf{r}_z\right| &= \left|\langle 2\cos\theta,2\sin\theta,0\rangle\right| = \sqrt{4\cos^2\theta+4\sin^2\theta} = 2 \\ \\ A(S) &= \int_0^{2\pi}\int_{-1}^3 2\,dz\,d\theta = 2\cdot 2\pi\cdot 4 = \color{answer}\mathbf{16\pi} \end{align*} \begin{align*} A(S) &= \iint_D\sqrt{1+\left( \frac{\partial z}{\partial x} \right)^2 + \left( \frac{\partial z}{\partial y} \right)^2}\,dA \\ &= \iint_D \sqrt{1+\tfrac{1}{4}(x^2+y^2)}\,dA = \int_0^{2\pi}\int_0^2 \sqrt{1+\tfrac{1}{4}r^2}r\,dr\,d\theta \\ &= 2\pi \int_0^2\sqrt{1+\tfrac{1}{4}r^2}r\,dr = 2\pi\bigg[ \frac{4}{3}(r^2+1)^{3/2} \bigg]_0^2 = \color{answer}\mathbf{\frac{8\pi}{3}(2\sqrt{2}-1)} \end{align*}