8. Use the divergence theorem to evaluate the flux integral F-ndS whereF(x.yz)= (x' +cotan“(y°z'), y' - e*²', z² +In(x- y), ñ is the outward unit normalto S, and S is the surface of the solid enclosed in the hemisphere z= Va? - x? - y²and the plane z=0.

Answered: 8. Use the divergence theorem to…

Calculus For The Life Sciences

2nd Edition

ISBN:9780321964038

Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.

Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.

Chapter8: Further Techniques And Applications Of Integration

Section8.3: Volume And Average Value

Problem 12E

See similar textbooks

Related questions

Question

8. Use the divergence theorem to evaluate the flux integral F-ndS where
F(x.yz)= (x' +cotan“(y°z'), y' - e*²', z² +In(x- y), ñ is the outward unit normal
to S, and S is the surface of the solid enclosed in the hemisphere z= Va? - x? - y²
and the plane z=0.

Expert Solution

Step 1

Given:

$F (x, y, z) = <x^{3} + c o \tan^{- 1} (y^{2} z^{3}), y^{3} - e^{x^{2} + z^{3}}, z^{3} + \ln (x - y)>$

$n$ is the outward unit normal to S, and S is the surface of the solid enclosed in the hemisphere

$z = \sqrt{a^{2} - x^{2} - y^{2}}$ and the plane $z = 0 .$

We have to evaluate the flux integral $\underset{s}{∯} \bar{F} . \bar{n} d S$ by using the divergence theorem

Step 2

The Divergence Theorem:

$\iint_{S} F . d S = \underset{E}{∭} div F d V$

Where S is a closed surface.

And E is the region inside that surface.

In this problem we have to calculate the flux of the integral which means we have to calculate

the surface integral but by using divergence theorem we will calculate the volume of the

given integral.

Now we will find div F:

$div F = \frac{\partial}{\partial x} (x^{3} + c o \tan^{- 1} (y^{2} z^{3})) + \frac{\partial}{\partial y} (y^{3} - e^{x^{2} + z^{3}}) + \frac{\partial}{\partial z} (z^{3} + \ln (x - y)) = 3 x^{2} + 3 y^{2} + 3 z^{2}$

Step 3

We have to calculate the integral $\underset{E}{∭} div F d V = \underset{E}{∭} div F d z d y d x$ :

The limit for z is:

$0 \leq z \leq \sqrt{a^{2} - x^{2} - y^{2}}$

and substitute

$x = r \cos θ y = r \sin θ$

We can define the region E as follows:

$E = \{(r, θ, z) | 0 \leq r \leq a, 0 \leq θ \leq 2 π, 0 \leq z \leq \sqrt{a^{2} - x^{2} - y^{2}}\}$

And Jacobian is:

$J = r$

Step by step

Solved in 5 steps

Knowledge Booster

$Triple Integral$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, advanced-math and related others by exploring similar questions and additional content below.

Recommended textbooks for you

Calculus For The Life Sciences

Calculus

ISBN:

9780321964038

Author:

GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.

Publisher:

Pearson Addison Wesley,

Calculus For The Life Sciences

Calculus

ISBN:

9780321964038

Author:

GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.

Publisher:

Pearson Addison Wesley,

SEE MORE TEXTBOOKS

8. Use the divergence theorem to evaluate the flux integral OF-nds where F(x,y.z)= (x' + cotan*(y`z'), y' - e*²', z' +In(x – y), ñ is the outward unit normal to S, and S is the surface of the solid enclosed in the hemisphere z = Ja² – x² - y? and the plane z=0.

8. Use the divergence theorem to evaluate the flux integral OF-nds where F(x,y.z)= (x' + cotan(y`z'), y' - e²', z' +In(x – y), ñ is the outward unit normal to S, and S is the surface of the solid enclosed in the hemisphere z = Ja² – x² - y? and the plane z=0.