Calculating Flux over the closed surface of a cylinder

[HRubss]

Homework Statement

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

Relevant Equations

$$\int \overrightarrow{V} \cdot \hat{n} da$$
$$\int\int ( \overrightarrow{V}\cdot\hat{n} )dS = \int \int \int (\nabla \cdot \overrightarrow{V}) dV$$

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. \oint \oint (\overrightarrow{V}\cdot\hat{n}) dS = \oint \oint (\overrightarrow{V} \cdot\hat{n}) d\theta dz
x = acos\thetay = asin\theta \overrightarrow{V} = <acos\theta, asin\theta,z>and \hat{n} =\frac{\partial r}{\partial \theta} \times \frac{\partial r}{\partial z}

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

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

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

 

[member 587159]

What is $r$ in the definition of $\hat{n}$? All integrals you calculated are correct, so the mistake did not happen there. I guess the mistake happens when you get to the expression of the first integral. Can't tell because you left out those calculations.

Note that $$\iiint 3dV = 3\iiint dV = 3 \mathrm{volume}(cylinder) = 3 \pi a^2b$$ so there you could have worked a little bit less ;).

 

[Orodruin]

You are forgetting the end-caps when doing the surface integral.

It is also not clear to me how you ended up with the $z$ in the integrand of the surface integral. The resulting term is quite clearly incorrect based on dimensional analysis.

 

[HRubss]

What is $r$ in the definition of $\hat{n}$? All integrals you calculated are correct, so the mistake did not happen there. I guess the mistake happens when you get to the expression of the first integral. Can't tell because you left out those calculations.

Note that $$\iiint 3dV = 3\iiint dV = 3 \mathrm{volume}(cylinder) = 3 \pi a^2b$$ so there you could have worked a little bit less ;).

Wow haha, this would've save me more time

You are forgetting the end-caps when doing the surface integral.

It is also not clear to me how you ended up with the $z$ in the integrand of the surface integral. The resulting term is quite clearly incorrect based on dimensional analysis.

My apologies, it should be \overrightarrow{r} = <acos\theta,bcos\theta , z> This is from the parametric description of a cylinder.
End caps? My calculus is a bit rusty, its been about 2 years since I've worked on vector calculus.

 

[Orodruin]

This is from the parametric description of a cylinder.

No, that is just the mantle area of the cylinder. The cylinder surface also consists of two discs - the end caps. Without them you do not have a closed surface.

Edit: Also consider my comments about the z term.

 

[HRubss]

No, that is just the mantle area of the cylinder. The cylinder surface also consists of two discs - the end caps. Without them you do not have a closed surface.

Oh! I see what you mean. Would I have to add two more flux bounded by a circle to the total flux?

 

[Orodruin]

Oh! I see what you mean. Would I have to add two more flux bounded by a circle to the total flux?

Discs, not circles. Discs are two-dimensional, circles one-dimensional.

Also, see the comment anout the z-term.

 

[HRubss]

Discs, not circles. Discs are two-dimensional, circles one-dimensional.

Also, see the comment anout the z-term.

For the mantel, the normal would be \hat{n} = \frac{<x,y,0>}{a} and for the disks, it would be \hat{n} = \frac{<0,0,z>}{b} From those two normal, I can take the dot product of normal with the vector and get the total flux? Sorry if I'm being slow...