# Calculate the double integral of \iint f(x,y,z) \text{d}S on the domain S for the given...

## Question:

Calculate the double integral of {eq}\iint f(x,y,z) \text{d}S {/eq} on the domain {eq}S {/eq} for the given surface {eq}G(r, \theta) = (r \cos(\theta), r \sin(\theta), \theta), \ 0 \leq r \leq 5, \ 0 \leq \theta \leq 2\pi {/eq}, and function {eq}f(x,y,z) = \sqrt{x^2+y^2} {/eq}.

## Calculating Double Integral:

We have to calculate the double integral for the given surface with the given function over the intervals.

Formula for calculating the surface integral on the domain is {eq}\ \displaystyle \iint_{D} f(x,y,z) \text{d}S =\int_{a}^{b}\int_{c}^{d}f(G(r, \theta ))\left \| G_{r} \times G_{\theta} \right \| d\theta dr {/eq}.

Where {eq}\displaystyle G_{r} {/eq} and {eq}\displaystyle G_{\theta} {/eq} are the partial derivatives of the surface. Then, {eq}\displaystyle G_{r} \times G_{\theta} {/eq} is the cross product of the partial derivatives.

Let us consider the given domain {eq}\displaystyle S {/eq} for the given surface {eq}\displaystyle G(r, \theta) = (r \cos(\theta), r \sin(\theta), \theta), \ 0 \leq r \leq 5, \ 0 \leq \theta \leq 2\pi {/eq}, and function {eq}\displaystyle f(x,y,z) = \sqrt{x^2+y^2} {/eq}.

Calculating the double integral of {eq}\displaystyle \iint f(x,y,z) \text{d}S {/eq}:

{eq}\begin{align*} \displaystyle G_{r}(r, \theta) &= (\cos \left(\theta \right), \sin \left(\theta \right), 0) \\ \displaystyle G_{\theta}(r, \theta) &= (-r\sin \left(\theta \right), r\cos \left(\theta \right), 1) \\ \displaystyle G_{r} \times G_{\theta} &=\begin{vmatrix} i & j & k\\ \cos \left(\theta \right) & \sin \left(\theta \right) & 0\\ -r\sin \left(\theta \right) & r\cos \left(\theta \right) & 1 \end{vmatrix} \\ \displaystyle &=((1)(\sin \left(\theta \right))-(r\cos \left(\theta \right))(0))\vec{i}-((1)(\cos \left(\theta \right))-(-r\sin \left(\theta \right))(0))\vec{j}+((r\cos \left(\theta \right))(\cos \left(\theta \right))-(-r\sin \left(\theta \right))(\sin \left(\theta \right)))\vec{k} \\ \displaystyle G_{r} \times G_{\theta} &= \sin \left(\theta \right) \vec{i} -\cos \left(\theta \right) \vec{j}+ r\vec{k} \\ \displaystyle \left \| G_{r} \times G_{\theta} \right \| &=\sqrt{(\sin \left(\theta \right))^{2}+(-\cos \left(\theta \right))^{2}+(r)^{2}} \\ \displaystyle \left \| G_{r} \times G_{\theta} \right \| &=\sqrt{r^2+1} \\ \displaystyle \iint f(x,y,z) \text{d}S &=\int_{a}^{b}\int_{c}^{d}f(G(r, \theta ))\left \| G_{r} \times G_{\theta} \right \| d\theta dr \\ \displaystyle &=\int_{0}^{5}\int_{0}^{2\pi}\left( \sqrt{(r \cos(\theta))^2+(r \sin(\theta))^2} \right)\left( \sqrt{r^2+1} \right) d\theta dr \\ \displaystyle &=\int_{0}^{5}\int_{0}^{2\pi} r \left( \sqrt{r^2+1} \right) d\theta dr \\ \displaystyle &=\int_{0}^{5} \left[ \sqrt{1+r^2}r\theta \right]_{0}^{2\pi} dr \\ \displaystyle &=\int_{0}^{5} \left[ \left( \sqrt{1+r^2}r(2\pi) \right)-\left( \sqrt{1+r^2}r(0) \right) \right] dr \\ \displaystyle &=\int_{0}^{5} \left[ 2\pi r\sqrt{r^2+1} \right] dr \\ \displaystyle &=2\pi \int_{0}^{5} r\sqrt{r^2+1} dr \\ \displaystyle \mathrm{Applying\:u-substitution:} \\ \displaystyle u &=r^2+1, \: du={(r^2+1)}' dr=2r dr, \: r dr=\frac{1}{2}du \\ \displaystyle &=2\pi \int_{0}^{5} \sqrt{u} \frac{1}{2}du \\ \displaystyle &=2\pi\cdot \frac{1}{2} \int_{0}^{5} u^{\frac{1}{2}} du \\ \displaystyle &=\pi \left[ \frac{2}{3}u^{\frac{3}{2}} \right]_{0}^{5} \\ \displaystyle \mathrm{Substituting, \:}\: u &= r^2+1 \\ \displaystyle &=\pi \left[ \frac{2}{3}(r^2+1)^{\frac{3}{2}} \right]_{0}^{5} \\ \displaystyle &=\pi \left[ \left( \frac{2}{3}((5)^2+1)^{\frac{3}{2}} \right)-\left( \frac{2}{3}((0)^2+1)^{\frac{3}{2}} \right) \right] \\ \displaystyle &=\pi \left[ \frac{52\sqrt{26}-2}{3} \right] \\ \displaystyle \iint f(x,y,z) \text{d}S&=275.56900 \end{align*} {/eq}

Answer for the double integral is {eq}\ \displaystyle \mathbf{\color{blue}{ \iint f(x,y,z) \text{d}S =275.57 }} {/eq}.