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I need to use the divergence theorem to evaluate the surface integral $$I = \int \int F\cdot n \, dS$$ where $F= x^3 i +y^3 j +z^3 k$ and $S$ is the surface of the cylinder $x^2+y^2 =4$ between $-1<z<1$

I know that first i take the divergence of $F$ which is $3r^2$.

Then in the coordinates

$$\int\int\int 3r^{3/2}\,dz\,dr\,d\theta$$

I am not sure what limits are though. $\theta$ would be $0$ to $2\pi$. I think $z$ would just be $-1$ to $1$. Am I on the right track?

Michael Hardy
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Jackson Hart
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1 Answers1

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If $F = x^3\hat{i}+y^3\hat{j}+z^3\hat{k}$, then $\nabla \cdot F = \dfrac{\partial}{\partial x}[x^3]+\dfrac{\partial}{\partial y}[y^3]+\dfrac{\partial}{\partial z}[z^3] = 3x^2+3y^2+3z^2$.

So by the divergence theorem, $\displaystyle\iint\limits_{S}F \cdot \hat{n}\,dS = \iiint\limits_{V}\nabla \cdot F \,dV = \iiint\limits_{V} 3(x^2+y^2+z^2)\,dx\,dy\,dx$.

Now convert this to cylindrical coordinates. You have the right bounds for $\theta$ and $z$. What about $r$?

Also, since $x^2+y^2 = r^2$, we have $3(x^2+y^2+z^2) = 3(r^2+z^2)$ not $3r^2$.

JimmyK4542
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