Let F = [x^2,0,z^2], and S the surface of the box abs(x) <= 1, abs(y) <= 3, 0 <= z <= 2. Evaluate the surface integral int int_S F*n dA by the divergence theorem.

Question

Question

Answers (1)

Relevant Questions

asked 2020-12-06

Use the Divergence Theorem to calculate the surface integral \(\displaystyle\int\int_{{S}}{F}·{d}{S}\), that is, calculate the flux of F across S.
\(\displaystyle{F}{\left({x},{y},{z}\right)}={\left({\cos{{\left({z}\right)}}}+{x}{y}^{{2}}\right)}{i}+{x}{e}^{{-{{z}}}}{j}+{\left({\sin{{\left({y}\right)}}}+{x}^{{2}}{z}\right)}{k}\)
S is the surface of the solid bounded by the paraboloid \(\displaystyle{z}={x}^{{2}}+{y}^{{2}}\) and the plane z = 9.

See answers (1)

asked 2021-01-31

Use the Divergence Theorem to calculate the surface integral \(\displaystyle\int\int_{{S}}{F}·{d}{S}\), that is, calculate the flux of F across S.
\(\displaystyle{F}{\left({x},{y},{z}\right)}={\left({x}^{{3}}+{y}^{{3}}\right)}{i}+{\left({y}^{{3}}+{z}^{{3}}\right)}{j}+{\left({z}^{{3}}+{x}^{{3}}\right)}{k}\), S is the sphere with center the origin and radius 2.

See answers (1)

asked 2021-03-11

Use the divergence theorem to evaluate \(\displaystyle\int\int_{{S}}{F}\cdot{N}{d}{S}\), where \(\displaystyle{F}{\left({x},{y},{z}\right)}={y}^{{2}}{z}{i}+{y}^{{3}}{j}+{x}{z}{k}\) and S is the boundary of the cube defined by \(\displaystyle-{5}\le{x},={5},-{5}\le{y}\le{5},{\quad\text{and}\quad}{0}\le{z}\le{10}\).

See answers (1)

asked 2021-01-19

Using the Divergence Theorem, evaluate \(\displaystyle\int\int_{{S}}{F}.{N}{d}{S}\), where \(\displaystyle{F}{\left({x},{y},{z}\right)}={\left({z}^{{3}}{i}-{x}^{{3}}{j}+{y}^{{3}}{k}\right)}\) and S is the sphere \(\displaystyle{x}^{{2}}+{y}^{{2}}+{z}^{{2}}={a}^{{2}}\), with outward unit normal vector N.

See answers (1)

asked 2020-12-28

Use the Divergence Theorem to calculate the surface integral F · dS, that is, calculate the flux of F across S.
\(\displaystyle{F}{\left({x},{y},{z}\right)}={3}{x}{y}^{{2}}{i}+{x}{e}^{{z}}{j}+{z}^{{3}}{k}\),
S is the surface of the solid bounded by the cylinder \(\displaystyle{y}^{{2}}+{z}^{{2}}={9}\) and the planes x = −3 and x = 1.

See answers (1)

asked 2021-03-09

Use Stokes' Theorem to evaluate \(\displaystyle\int\int_{{S}}{C}{U}{R}{L}{f}\cdot{d}{S}\).
\(\displaystyle{F}{\left({x},{y},{z}\right)}={x}^{{2}}{y}^{{3}}{z}{i}+{\sin{{\left({x}{y}{z}\right)}}}{j}+{x}{y}{z}{k}\),
S is the part of the cone \(\displaystyle{y}^{{2}}={x}^{{2}}+{z}^{{2}}\) that lies between the planes y = 0 and y = 2, oriented in the direction of the positive y-axis.

See answers (1)

asked 2021-02-05

Let C be the ellipse contained in the xy plane whose equation is \(\displaystyle{4}{x}^{{2}}+{y}^{{2}}={4}\), oriented clockwise. The force field F described by \(\displaystyle{F}{\left({x},{y},{z}\right)}={x}^{{2}}{i}+{2}{x}{j}+{z}^{{2}}{k}\), moves a particle along C in the same direction as the curve orientation, performing a W job. C as the surface boundary S: \(\displaystyle{z}={4}-{4}{x}^{{2}}-{y}^{{2}},{z}\ge{0}\) (with ascending orientation, that is, the component in the z direction equal to 1) and assuming \(\displaystyle\pi={3.14}\), we can state what:
a) It is not necessary to apply Stokes' Theorem, as C is a closed curve and therefore W = 0.
b) Inverting the orientation of the surface S, we can apply Stokes' Theorem and conclude that W = 12.56.
c) We can apply Stokes' Theorem and conclude that W = 6.28
d) We can apply Stokes' Theorem and conclude that W = 12.56.

See answers (1)

asked 2021-01-10

Let \(\displaystyle{F}\vec{\to}{r}={<}{x},{y},{z}{>}\) and use the Divergence Theorem to calculate the (nonzero) volume of some solid in IR3 by calculating a surface integral. (You can pick the solid).

See answers (1)

asked 2021-01-31

Flux integrals Compute the outward flux of the following vector fields across the given surfaces S. You should decide which integral of the Divergence Theorem to use.
\(\displaystyle{F}={\left\langle{x}{\sin{{y}}},-{\cos{{y}}},{z}{\sin{{y}}}\right\rangle}\) , S is the boundary of the region bounded by the planes x = 1, y = 0, \(\displaystyle{y}=\frac{\pi}{{2}},{z}={0}\), and z = x.

See answers (1)

asked 2021-03-08

z = x Let be the curve of intersection of the cylinder \(\displaystyle{x}^{{2}}+{y}^{{2}}={1}\) and the plane , oriented positively when viewed from above . Let S be the inside of this curve , oriented with upward -pointing normal . Use Stokes ' Theorem to evaluate \(\displaystyle\int{S}{c}{u}{r}{l}{F}\cdot{d}{S}{\quad\text{if}\quad}{F}={y}{i}+{z}{j}+{2}{x}{k}\).

See answers (1)

Answer 1

Step 1
The divergence of the function F is:
\(\displaystyle\nabla\cdot{F}=\frac{{\partial{P}}}{{\partial{x}}}+\frac{{\partial{Q}}}{{\partial{y}}}+\frac{{\partial{R}}}{{\partial{z}}}\)
\(\displaystyle\div{F}=\frac{\partial}{{\partial{x}}}{\left({x}^{{2}}\right)}+\frac{\partial}{{\partial{y}}}{\left({0}\right)}+\frac{\partial}{{\partial{z}}}{\left({z}^{{2}}\right)}\)
=2x+2z
Step 2
By the divergence theorem convert the surface integral into a triple integral:
\(\displaystyle\int\int_{{S}}{P}{\left.{d}{y}\right.}{\left.{d}{z}\right.}+{Q}{\left.{d}{x}\right.}{\left.{d}{z}\right.}+{R}{\left.{d}{x}\right.}{\left.{d}{y}\right.}=\int\int_{{V}}\int{\left(\frac{{\partial{P}}}{{\partial{x}}}+\frac{{\partial{Q}}}{{\partial{y}}}+\frac{{\partial{R}}}{{\partial{z}}}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}{\left.{d}{x}\right.}\)
\(\displaystyle=\int\int_{{V}}\int{\left({2}{x}+{2}{z}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}{\left.{d}{z}\right.}\)
Step 3
\(\displaystyle\int\int_{{V}}\int{\left({2}{x}+{2}{z}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}{\left.{d}{z}\right.}={\int_{{-{{1}}}}^{{1}}}{\left.{d}{x}\right.}{\int_{{3}}^{{3}}}{\left.{d}{y}\right.}{\int_{{0}}^{{2}}}{\left({2}{x}+{2}{z}\right)}{\left.{d}{z}\right.}\)
\(\displaystyle={2}{\int_{{1}}^{{1}}}{\left.{d}{x}\right.}{\int_{{3}}^{{3}}}{\left.{d}{y}\right.}{\in_{{0}}^{{2}}}{\left({x}+{z}\right)}{\left.{d}{z}\right.}\)
\(\displaystyle={2}{\int_{{1}}^{{1}}}{\left.{d}{x}\right.}{\int_{{3}}^{{3}}}{\left.{d}{y}\right.}{\left[{\left({x}{z}+\frac{{z}^{{2}}}{{2}}\right)}{{\mid}_{{0}}^{{2}}}\right]}\)
\(\displaystyle={2}{\int_{{1}}^{{1}}}{\left.{d}{x}\right.}{\int_{{3}}^{{3}}}{\left({2}{x}+{2}\right)}{\left.{d}{y}\right.}\)
\(\displaystyle={4}{\int_{{1}}^{{1}}}{\left.{d}{x}\right.}{\left[{\left({x}{y}+{y}\right)}{{\mid}_{{-{{3}}}}^{{3}}}\right]}\)
\(\displaystyle={4}{\int_{{1}}^{{1}}}{\left.{d}{x}\right.}{\left[{\left({3}{x}+{3}\right)}-{\left(-{3}{x}-{3}\right)}\right]}\)
\(\displaystyle={4}{\int_{{1}}^{{1}}}{\left.{d}{x}\right.}{\left[{\left({6}{x}+{6}\right)}\right]}\)
\(\displaystyle={24}{\int_{{1}}^{{1}}}{\left.{d}{x}\right.}\)
\(\displaystyle={24}{{\left[{x}\right]}_{{-{{1}}}}^{{1}}}\)
=48

Didn’t find what you are looking for?

Let F = [x^2,0,z^2], and S the surface of the box abs(x) <= 1, abs(y) <= 3, 0 <= z <= 2. Evaluate the surface integral int int_S F*n dA by the divergence theorem.

Answers (1)

Relevant Questions