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Q. 34

Expert-verified
Found in: Page 1119

### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861

# , where S is the surface given by $\mathbit{r}\left(u,v\right)=2u\mathbf{i}+\left(\mathrm{u}+\mathrm{v}\right)\mathbf{j}+\left(2\mathrm{u}-2\mathrm{v}\right)\mathbf{k}$ for $3\le u\le 7$ and $2\le v\le 10$.

The required integral over the surface S is $24\mathrm{ln}\left(\frac{\mathrm{ln}\left(28\right)}{\mathrm{ln}\left(12\right)}\right)$.

See the step by step solution

## Step 1. Given information.

Consider the given question,

## Step 2. Find the integral of fx,y,z over a smooth surface S.

If a surface S is parametrized by $r\left(u,v\right)$ for $\left(u,v\right)\in D$, then the integral of $f\left(x,y,z\right)$ over a smooth surface S is given below,

${\int }_{S}f\left(x,y,z\right)dS=\iint {}_{D}f\left(x\left(u,v\right),y\left(u,v\right),z\left(u,v\right)\right)||{\mathbit{r}}_{u}×{\mathbit{r}}_{w}||dA......\left(i\right)$

Now, find ${\mathbit{r}}_{\mathbf{u}},{\mathbit{r}}_{v}$,

role="math" localid="1650303986567" ${\mathbit{r}}_{u}=\frac{\partial }{\partial u}\mathbit{r}\left(u,v\right)\phantom{\rule{0ex}{0ex}}=\frac{\partial }{\partial u}\left(2u\mathbf{i}+\left(\mathrm{u}+\mathrm{v}\right)\mathbf{j}+\left(2\mathrm{u}-2\mathrm{v}\right)\mathbf{k}\right)\phantom{\rule{0ex}{0ex}}=2\mathbit{i}+\mathbit{j}+2\mathbit{k}\phantom{\rule{0ex}{0ex}}{r}_{v}=\frac{\partial }{\partial v}\mathbit{r}\left(u,v\right)\phantom{\rule{0ex}{0ex}}=\frac{\partial }{\partial v}\left(2u\mathbf{i}+\left(\mathrm{u}+\mathrm{v}\right)\mathbf{j}+\left(2\mathrm{u}-2\mathrm{v}\right)\mathbf{k}\right)\phantom{\rule{0ex}{0ex}}=0\mathbit{i}+\mathbit{j}-2\mathbit{k}$

## Step 3. Find ru×rv.

On solving $||{\mathbit{r}}_{u}×{\mathbit{r}}_{\mathbf{v}}||$,

${\mathbit{r}}_{u}×{\mathbit{r}}_{\mathrm{v}}=\left(2\mathbit{i}+\mathbit{j}+2\mathbit{k}\right)×\left(0\mathbit{i}+\mathbit{j}-2\mathbit{k}\right)\phantom{\rule{0ex}{0ex}}=\left|\begin{array}{ccc}\mathbit{i}& \mathbit{j}& \mathbit{k}\\ 2& 1& 2\\ 0& 1& -2\end{array}\right|\phantom{\rule{0ex}{0ex}}=-4\mathbit{i}+4\mathbit{j}+2\mathbit{k}$

Then,$||{\mathbit{r}}_{u}×{\mathbit{r}}_{\mathrm{v}}||=||-4\mathbf{i}+4\mathbf{j}+2\mathbf{k}||\phantom{\rule{0ex}{0ex}}=\sqrt{{\left(-4\right)}^{2}+{4}^{2}+{2}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{36}\phantom{\rule{0ex}{0ex}}=6$

Consider the following function, localid="1650304882776" $f\left(x,y,z\right)$,

localid="1650304885354" $f\left(x\left(u,v\right),y\left(u,v\right),z\left(u,v\right)\right)=\frac{1}{2u\mathrm{ln}\left(\left(2u-2v\right)+2\left(u+v\right)\right)}\phantom{\rule{0ex}{0ex}}=\frac{1}{2u\mathrm{ln}\left(4u\right)}$

## Step 4. Consider the given function.

Substitute the values in equation (i),

$D=\left\{\left(u,v\right)\overline{)3\le u\le 7,2\le u\le 10}\right\}$

Take the integral of f over the surface S,

$\begin{array}{r}{\int }_{S} f\left(x,y,z\right)dS={\iint }_{D} f\left(x\left(u,v\right),y\left(u,v\right),z\left(u,v\right)\right)∥{\mathbf{r}}_{u}×{\mathbf{r}}_{v}∥dA\\ ={\int }_{2}^{10} {\int }_{3}^{7} \left[\frac{1}{2u\mathrm{ln}\left(4u\right)}\right)\left(6\right)dudv\\ =3{\int }_{2}^{10} {\int }_{3}^{7} \frac{1}{u\mathrm{ln}\left(4u\right)}dudv\\ =3{\int }_{2}^{10} \left[{\int }_{3}^{7} \frac{1}{u\mathrm{ln}\left(4u\right)}du\right]dv\\ =3{\int }_{2}^{10} \left[\mathrm{ln}\left(\mathrm{ln}\left(4u\right)\right){\right]}_{3}^{7}dv\end{array}$

## Step 5. Continue solving the above equation.

On solving the above equation,

$\begin{array}{r}{\int }_{S} f\left(x,y,z\right)dS=3{\int }_{2}^{10} \left[\mathrm{ln}\left(\mathrm{ln}\left(28\right)\right)-\mathrm{ln}\left(\mathrm{ln}\left(12\right)\right)\right]dv\\ =3{\int }_{2}^{10} \mathrm{ln}\left(\frac{\mathrm{ln}\left(28\right)}{\mathrm{ln}\left(12\right)}\right)dv\\ =3\mathrm{ln}\left(\frac{\mathrm{ln}\left(28\right)}{\mathrm{ln}\left(12\right)}\right){\int }_{2}^{10} dv\\ =3\mathrm{ln}\left(\frac{\mathrm{ln}\left(28\right)}{\mathrm{ln}\left(12\right)}\right)\left[v{\right]}_{2}^{10}\\ =3\mathrm{ln}\left(\frac{\mathrm{ln}\left(28\right)}{\mathrm{ln}\left(12\right)}\right)\left[10-2\right]\\ =24\mathrm{ln}\left(\frac{\mathrm{ln}\left(28\right)}{\mathrm{ln}\left(12\right)}\right)\end{array}$