Surface integral problem using spherical coordinates

Question

Hi, for this question I used spherical coordinates. $x=\cos(\theta), y=\sin(\theta)$, $z=t$, $ds=d(\theta)dt$ But when I substitute into the equation I get $12\pi$ instead of $24\pi$(the answer). My limits for the first integration was $0$ to $1$ and the second integral was from $0$ to $2\pi$. Did i do something wrong here?.

Answer 1

You have the parametric surface $r(\theta, t) = (2\cos(\theta),2\sin(\theta),t)$ where the parameters $(\theta,t) \in [0,2\pi] \times [0,1]$. Also, $dS = |r_{\theta} \times r_t|dA$, where the subscript indicates taking partial derivatives with respect to that variable. So Calculating $|r_{\theta} \times r_t| = 2$. So we have

\begin{align*} \iint_S (5+2z)dS & = \int_{0}^{2\pi}\int_{0}^1 (5+2t) |r_{\theta}\times r_t|dtd\theta \\ & = 2 \int_0^{2\pi}\int_0^1 (5+2t)dtd\theta \\ & = 2\int_0^{2\pi}[5t+t^2]_0^1 d\theta \\ & = 2\int_0^{2\pi}6d\theta \\& = 2[6\theta]_0^{2\pi} \\ & = 24\pi \end{align*}

\begin{align*} \frac{\partial}{\partial \theta}r(\theta,t) & = (-2\sin(\theta), 2\cos(\theta),0) \\ \frac{\partial}{\partial t}r(\theta,t) & = (0,0,1) \end{align*}

So we have that

\begin{align*} r_{\theta} \times r_{t} & = \left|\begin{matrix} i & j & k \\ -2\sin(\theta) & 2\cos(\theta) & 0 \\ 0 & 0 & 1 \end{matrix}\right| \\& = \left|\begin{matrix} 2\cos(\theta) & 0 \\ 0 & 1 \end{matrix}\right|i - \left|\begin{matrix} -2\sin(\theta) & 0 \\ 0 & 1\end{matrix}\right|j + \left|\begin{matrix} -2\sin(\theta) & 2\cos(\theta) \\ 0 & 0 \end{matrix}\right|k \\ & = (2\cos(\theta), -(-2\sin(\theta)), 0) \\ & = (2\cos(\theta), 2\sin(\theta), 0) \end{align*}

So you have that $|r_{\theta} \times r_{t}| = \sqrt{(2\cos(\theta))^2 + (2\sin(\theta))^2} = \sqrt{4\cos^2(\theta) + 4\sin^2(\theta)} = 2$

Answer 2

An elementary solution

The surface over which we integrate is a cylinder of height $1$ and radius $2$ as shown in the figure below.

The surface area of the red ribbon is $4\pi\Delta z$. The function to be integrated over the surface is $\approx$ a constant around the ribbon; it depends only on $z$. The product of the red surface area and the value of the function at $z$ is $$\approx 4\pi(5+2z)\Delta z.$$

The surface integral is then

$$4\pi\int_0^1(5+2z)\ dz=24\pi.$$