Let $S=\{(x,y,z)\in \mathbb{R} : x^2+y^2+z^2 = 11\}$ and $F(x,y,z)=(2\sqrt{11}x,y^2,z^2)$. Calculate $$\int\int_SF(x,y,z)\;dS$$

I've calculated S parameterized, whose expression is:

$$S(x,y) = (x,y,\sqrt{11-x^2-y^2})$$

Hence, $$r_x=\frac{\partial S}{\partial x}=(1,0,\frac{-x}{\sqrt{11-x^2-y^2}})$$ $$r_y=\frac{\partial S}{\partial y}=(0,1,\frac{-y}{\sqrt{11-x^2-y^2}})$$ $$||r_x \times r_y|| = \frac{\sqrt{11}}{\sqrt{11-x^2-y^2}}$$

So $dS = \frac{\sqrt{11}}{\sqrt{11-x^2-y^2}}dxdy$ and then

$$\int\int_SF(x,y,z)\;dS = \int\int_SF(x,y,z)\;\frac{\sqrt{11}}{\sqrt{11-x^2-y^2}}dxdy$$

But then I don't know how to continue with this procedure.

Another idea I have is to establish the following relationship:

$$\int\int_SF\;dS= \int\int\int_V \nabla F\;dV$$

Any hint?

Thanks in advance.

  • 1
    $\begingroup$ Since $ S $ is compact and $F$ contains no singularites you can calculate it with ease with the divergence theorem that you posted. Hint: Use spherical coordinates to parameterize the volume contained by $ S $ and don't forget the Jacobian. =) $\endgroup$
    – Oskar
    Commented Apr 29, 2020 at 13:18
  • $\begingroup$ Thank you @Oskar. What do you mean by the Jacobian? $\endgroup$
    – user9867
    Commented Apr 29, 2020 at 13:21
  • $\begingroup$ I mean, what functions do I have to introduce in the Jacobian? $\endgroup$
    – user9867
    Commented Apr 29, 2020 at 13:29
  • $\begingroup$ I think @Oskar is referring to the Jacobian for the transformation to spherical coordinates. That is,$$\mathrm dV=\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi$$ $\endgroup$
    – user170231
    Commented Apr 29, 2020 at 16:15

1 Answer 1


Parameterizing $S$ in spherical coordinates makes this easier, though not by much.

$$\vec r(u,v)=\sqrt{11}(\cos u\sin v,\sin u\sin v,\cos v)$$

with $0\le u\le2\pi$ and $0\le v\le\pi$.

Then take the normal vector to $S$ to be

$$\vec r_v\times\vec r_u=11(\cos u\sin^2v,\sin u\sin^2v,\cos v\sin v)$$

where this order of partial derivatives is taken to make the normal vector point outward, away from the origin.

Then the integral is

$$\begin{align*} \iint_S\vec F(x,y,z)\cdot\mathrm d\vec S&=\int_0^\pi\int_0^{2\pi}\vec F(\vec r(u,v))\cdot(\vec r_v\times\vec r_u)\,\mathrm du\,\mathrm dv\\[1ex] &=\int_0^\pi\int_0^{2\pi}(22\cos u\sin v,11\sin^2u\sin^2v,11\cos^2v)\cdot(\vec r_v\times\vec r_u)\,\mathrm du\,\mathrm dv \end{align*}$$

The resulting integral is a bit complicated, but doable by hand. Of course, resorting to the divergence theorem makes this significantly easier to compute:

$$\begin{align*} \iint_S\vec F\cdot\mathrm d\vec S&=\iiint_{x^2+y^2+z^2\le11}2\sqrt{11}+2y+2z\,\mathrm dV\\[1ex] &=\int_0^\pi\int_0^{2\pi}\int_0^{\sqrt{11}}(\sqrt{11}+2\rho\sin\theta\sin\varphi+2\rho\cos\varphi)\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi \end{align*}$$

Either way, you would get


  • $\begingroup$ Thank you very much, @user170231. Now I have all muy doubts clear. $\endgroup$
    – user9867
    Commented Apr 29, 2020 at 21:48

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