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Calculate the flux JF ds through the surface of the sphere in the figure of the vector field: F(R,0,0) =(_nkz J8R 8 cos(e) @ 4R 6 The field is written in spherical coordinate system: The radius Ro is equal to 7. The sphere is centered at x=0, y=0 and z=0 and assume that the normal at the surface is oriented outward:

Ro

Figure: The surface of the sphere used to do the flux or the surface integral of the vector field.

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03:31

Flux across a sphere Consider the radial field $\mathbf{F}=\langle x, y, z\rangle$ and let $S$ be the sphere of radius $a$ centered at the origin. Compute the outward flux of $\mathbf{F}$ across $S$ using the representation $z=\pm \sqrt{a^{2}-x^{2}-y^{2}}$ for the sphere (either symmetry or two surfaces must be used).

00:10

Use the Divergence Theorem to calculate the surface integral $ \iint_S \textbf{F} \cdot d\textbf{S} $; that is, calculate the flux of $ \textbf{F} $ across $ S $.

$ \textbf{F} = | \textbf{r} |^2 \textbf{r} $, where $ \textbf{r} = x \, \textbf{i} + y \, \textbf{j} + z \, \textbf{k} $,$ S $ is the sphere with radius $ R $ and center the origin

03:44

Surface integrals of vector fields Find the flux of the following vector fields across the given surface with the specified orientation. You may use either an explicit or parametric description of the surface.$\mathbf{F}=\mathbf{r} /|\mathbf{r}|^{3}$ across the sphere of radius $a$ centered at the origin, where $\mathbf{r}=\langle x, y, z\rangle$; the normal vectors point outward.

02:37

Calculate the flux of the vector field $\mathbf{F}=2 x y \mathbf{i}-y^{2} \mathbf{j}+\mathbf{k}$ through the surface $\mathcal{S}$ in Figure $18 .$ Hint: Apply the Divergence Theorem to the closed surface consisting of $\mathcal{S}$ and the unit disk.

Transcript

In this question we have given, if r theta is equal to 1 divide by 4 pi r square multiplied by 8 r cap plus 8 cos theta 5 plus 4 r theta cap, and we have given a sphere, is centered at x, equal to 0 y equal To 0 and z equal to 0 point and we have to calculate the flux, seethe solution sphere is equal to r square sine pi d. Pi d theta is equal to d s: f, dot d s is equal to 8 divided by 4 pi r square multiplied by r square sine pi d, fi d, theta from here r square r square, would cancel our flux. Our flux is 0 to 2 pi integration, 0 to 2 pi integration, 0 to pi 2 divide by pi…

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