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HRubss

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- Homework Statement
- Calculate the flux of [tex]\overrightarrow{V} = x\hat{i} + y\hat{j} + z\hat{k}[/tex][tex]\int \overrightarrow{V} \cdot \hat{n} da [/tex] where the integral is to be taken over the closed surface of a cylinder [tex]x^2 + y^2 = a^2[/tex] which is bounded by the place z = 0 and z = b

- Relevant Equations
- [tex]\int \overrightarrow{V} \cdot \hat{n} da [/tex]

[tex]\int\int ( \overrightarrow{V}\cdot\hat{n} )dS = \int \int \int (\nabla \cdot \overrightarrow{V}) dV[/tex]

I wanted to check my answer because I'm getting two different answers with the use of the the Divergence theorem. For the left part of the equation, I converted it so that I can evaluate the integral in polar coordinates. [tex]\oint \oint (\overrightarrow{V}\cdot\hat{n}) dS = \oint \oint (\overrightarrow{V} \cdot\hat{n}) d\theta dz[/tex]

[tex]x = acos\theta[/tex][tex]y = asin\theta[/tex] [tex]\overrightarrow{V} = <acos\theta, asin\theta,z>[/tex]and [tex]\hat{n} =\frac{\partial r}{\partial \theta} \times \frac{\partial r}{\partial z} [/tex]

After applying the cross product and the dot product, I end up with this integral to evaluate. [tex]\int_{0}^{b}\int_{0}^{2\pi}(a^2cos^2\theta + a^2sin^2\theta + z )d\theta dz[/tex] which gives me [tex](2a^2+b)b\pi[/tex]

When I do the right side of the equation, taking the divergence of V results in this integral. [tex]\int \int \int(3)dV[/tex] and converting it to polar via [tex]dV = rdrdzd\theta[/tex] results in [tex]\int_{0}^{2\pi}\int_{0}^{b}\int_{0}^{a}(3r)drdzd\theta [/tex] I finally end up with [tex]3\pi a^2b[/tex]

Is my arithmetic off or am I doing something wrong? Thank you for any responses!

[tex]x = acos\theta[/tex][tex]y = asin\theta[/tex] [tex]\overrightarrow{V} = <acos\theta, asin\theta,z>[/tex]and [tex]\hat{n} =\frac{\partial r}{\partial \theta} \times \frac{\partial r}{\partial z} [/tex]

After applying the cross product and the dot product, I end up with this integral to evaluate. [tex]\int_{0}^{b}\int_{0}^{2\pi}(a^2cos^2\theta + a^2sin^2\theta + z )d\theta dz[/tex] which gives me [tex](2a^2+b)b\pi[/tex]

When I do the right side of the equation, taking the divergence of V results in this integral. [tex]\int \int \int(3)dV[/tex] and converting it to polar via [tex]dV = rdrdzd\theta[/tex] results in [tex]\int_{0}^{2\pi}\int_{0}^{b}\int_{0}^{a}(3r)drdzd\theta [/tex] I finally end up with [tex]3\pi a^2b[/tex]

Is my arithmetic off or am I doing something wrong? Thank you for any responses!

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