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Answers without the blur. Sign up and see all textbooks for free! Q. 40

Expert-verified Found in: Page 1120 ### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861 # $\mathbf{F}\left(x,y,z\right)=\mathbf{i}+\mathbf{j}+\mathbf{k}$, where S is the lower half of the unit sphere, with n pointing outwards.

The required flux of the vector field through the surface S is $2\mathrm{\pi }$.

See the step by step solution

## Step 1. Given information.

Consider the given question,

$\mathbf{F}\left(x,y,z\right)=\mathbf{i}+\mathbf{j}+\mathbf{k}$

## Step 2. Find the Flux of Fx,y,zthough an oriented surface S.

If a surface S is the graph of $z=z\left(x,y\right)$, then the Flux of $\mathbit{F}\left(x,y,z\right)$ through S is given below,

$\begin{array}{r}{\int }_{S} \mathbf{F}\left(x,y,z\right)\cdot \mathbf{n}dS={\iint }_{D} \left(\mathbf{F}\left(x,y,z\right)\cdot \mathbf{n}\right)∥{z}_{x}×{z}_{y}∥dA\\ ={\iint }_{D} \left(\mathbf{F}\left(x,y,z\right)\cdot \mathbf{n}\right)\sqrt{{\left(\frac{\mathrm{\partial }z}{\mathrm{\partial }x}\right)}^{2}+{\left(\frac{\mathrm{\partial }z}{\mathrm{\partial }y}\right)}^{2}+1}dA\dots ...\left(i\right)\end{array}$

Here, the surface S is the lower half of the unit sphere, so its equation will be,

$z=-\sqrt{1-{x}^{2}-{y}^{2}}$

Now first find $\frac{\mathrm{\partial }z}{\mathrm{\partial }x}$. The first partial derivates of z are given below,

$\frac{\mathrm{\partial }z}{\mathrm{\partial }x}=\frac{\mathrm{\partial }}{\mathrm{\partial }x}\left(-\sqrt{1-{x}^{2}-{y}^{2}}\right)\phantom{\rule{0ex}{0ex}}=\frac{x}{\sqrt{1-{x}^{2}-{y}^{2}}}$

## Step 3. Find the value of ∂z∂x2+∂z∂y2+1.

On finding the value of $\frac{\mathrm{\partial }z}{\mathrm{\partial }\mathrm{y}}$,

role="math" localid="1650342662824" $\frac{\mathrm{\partial }z}{\mathrm{\partial }\mathrm{y}}=\frac{\mathrm{\partial }}{\mathrm{\partial }\mathrm{y}}\left(-\sqrt{1-{x}^{2}-{y}^{2}}\right)\phantom{\rule{0ex}{0ex}}=\frac{y}{\sqrt{1-{x}^{2}-{y}^{2}}}$

Then,

$\begin{array}{r}\sqrt{{\left(\frac{\mathrm{\partial }z}{\mathrm{\partial }x}\right)}^{2}+{\left(\frac{\mathrm{\partial }z}{\mathrm{\partial }y}\right)}^{2}+1}=\sqrt{{\left(\frac{x}{\sqrt{1-{x}^{2}-{y}^{2}}}\right)}^{2}+{\left(\frac{y}{\sqrt{1-{x}^{2}-{y}^{2}}}\right)}^{2}+1}\\ =\sqrt{\frac{{x}^{2}}{1-{x}^{2}-{y}^{2}}+\frac{{y}^{2}}{1-{x}^{2}-{y}^{2}}+1}\\ =\frac{\sqrt{1-{x}^{2}-{y}^{2}}}{\sqrt{1}}\end{array}$

## Step 4. Find the choice of n.

The choice of n should have pointing upwards. Then the following vector, is normal to the surface.

Now, for $z=-\sqrt{1-{x}^{2}-{y}^{2}}$, gives the following vector perpendicular to this surface,

Then the desired normal vector will be,

## Step 5. Find the value of F(x,y,z)⋅n.

The value of $\mathbf{F}\left(x,y,z\right)\cdot \mathbf{n}$ will be,

## Step 6. Find the polar coordinates.

Substituting these values in equation (i),

$\begin{array}{r}{\int }_{S} \mathbf{F}\left(x,y,z\right)\cdot \mathbf{n}dS={\iint }_{D} \left(\mathbf{F}\left(x,y,z\right)\cdot \mathbf{n}\right)\left(\sqrt{{\left(\frac{\mathrm{\partial }z}{\mathrm{\partial }x}\right)}^{2}+{\left(\frac{\mathrm{\partial }z}{\mathrm{\partial }y}\right)}^{2}+1}\right)dA\\ ={\iint }_{D} \left(\sqrt{1-{x}^{2}-{y}^{2}}\left(1-\frac{x+y}{\sqrt{1-{x}^{2}-{y}^{2}}}\right)\right)\left(\frac{1}{\sqrt{1-{x}^{2}-{y}^{2}}}\right)dA\\ ={\iint }_{D} \left(1-\frac{x+y}{\sqrt{1-{x}^{2}-{y}^{2}}}\right)dA\dots \dots .\left(ii\right)\end{array}$

The surface S is the lower half of the unit sphere, so the region of integration D is the unit disk in the xy-plane and centered at the origin.

Hence, in polar coordinates, the region of integration will be,

$D=\left\{\left(r,\theta \right)\mid 0\le r\le 1,0\le \theta \le 2\pi \right\}$

In this case,

localid="1650343222895" $\begin{array}{r}x=r\mathrm{cos}\theta ,\\ y=r\mathrm{sin}\theta \end{array}$

## Step 7. Using equation (ii), write the flux of the vector field through the surface S.

Using equation (ii), the flux of the vector field through the surface S,

$\begin{array}{r}{\int }_{S} \mathbf{F}\left(x,y,z\right)\cdot \mathbf{n}dS={\iint }_{D} \left(1-\frac{x+y}{\sqrt{1-{x}^{2}-{y}^{2}}}\right)dA\\ ={\int }_{0}^{2\pi } {\int }_{0}^{1} \left(1-\frac{r\mathrm{cos}\theta +r\mathrm{sin}\theta }{\sqrt{1-{r}^{2}}}\right)rdrd\theta \\ ={\int }_{0}^{2\pi } \left[{\int }_{0}^{1} 1dr-\left(\mathrm{cos}\theta +\mathrm{sin}\theta \right){\int }_{0}^{1} \frac{{r}^{2}}{\sqrt{1-{r}^{2}}}dr\right]d\theta \\ ={\int }_{0}^{2\pi } \left[\left[1-0\right]-\left(\mathrm{cos}\theta +\mathrm{sin}\theta \right)\left[\frac{1}{2}\left(-1\sqrt{1-{1}^{2}}+{\mathrm{sin}}^{-1}1\right)-\frac{1}{2}\left(-0\sqrt{1-{0}^{2}}+{\mathrm{sin}}^{-1}0\right)\right]\right]d\theta \\ \begin{array}{r}\begin{array}{r}={\int }_{0}^{2\pi } \left[1-\frac{\pi }{4}\left(\mathrm{cos}\theta +\mathrm{sin}\theta \right)\right]d\theta \\ ={\left[\theta -\frac{\pi }{4}\left(\mathrm{sin}\theta -\mathrm{cos}\theta \right)\right]}_{0}^{2\pi }\\ =\left(2\pi -\frac{\pi }{4}\left(\mathrm{sin}2\pi -\mathrm{cos}2\pi \right)-\left(0-\frac{\pi }{4}\left(\mathrm{sin}0-\mathrm{cos}0\right)\right)\end{array}\\ =2\pi \end{array}\end{array}$ ### Want to see more solutions like these? 