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This was a question a students had asked me earlier today regarding surface area.

Find the surface area of the hemisphere $x^2+y^2+z^2 = 4$ bounded below by $z=1$.

I decided to approach this problem using spherical coordinates and got the following

\begin{eqnarray} \int_{0}^{2\pi}\int_{0}^{\pi/3}4\sin\phi d\phi d\theta & = & 4\int_{0}^{2\pi}d\theta\int_{0}^{\pi/3}\sin\phi d\phi\\ & = & 8\pi\cos\phi|_{\pi/3}^{0}\\ & = & 8\pi(1-\frac{1}{2}) = 4\pi \end{eqnarray}

I also solved this problem using single variable calculus as follows. I can represent the sphere as a circle of $h^2+z^2 =4$. Thus we have the following:

\begin{eqnarray} SA = \int_a^b2\pi f(z)ds & = & 2\pi\int_1^2\sqrt{4-z^2}\sqrt{1+\frac{z^2}{4-z^2}}dz\\ & = & 2\pi\int_1^2\sqrt{4-z^2}\sqrt{\frac{4}{4-z^2}} dz\\ & = & 2\pi\int_1^22dz=4\pi z|_1^2=4\pi \end{eqnarray}

As you see, I got the same answer for both approaches.

The student, and a few others, comes back later during the day and tells me the answer I got was incorrect. He does not tell me what the professor got, he just told me it was wrong. I asked myself "why?" Is there something I missed?

Thanks in advance for any feedback.

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  • $\begingroup$ Maybe there is something wrong with the text of the problem. The surface whose area has been calculated is not a hemisphere but a spherical cap. $\endgroup$ – Christian Blatter Nov 19 '14 at 10:52
  • $\begingroup$ Git Gud Could u please help me with this vector calculus question as well. I really need help in this. Thanks $\endgroup$ – ys wong Nov 23 '14 at 7:38
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I don't see what you did wrong here...

To me, to solve these types of problems you have to think geometrically--there isn't going to be some way to do it just from a knowledge of multivariable calculus.

First, how to find the total surface area of the sphere--that will help. You need to break the sphere up into circles stacked on top of each other, then find the $dA$:

$$ dA = 2\pi r_{\phi} h $$

$h$ is easy to find, it's just $rd\phi$, $r_\phi$ is the radius at the given azimuth: $r_\phi = r\sin(\phi)$ which gives:

$$ dA = 2\pi r^2\sin(\phi)d\phi\\ A = \int dA = 2\pi r^2\left.\int_{0}^{\pi}\sin(\phi)d\phi = -2\pi r^2\cos(\phi)\right|_0^\pi = 4\pi r^2 $$

So the correct integral should be:L

$$ A = 2\pi r^2\left.\int_{0}^{\phi_0} \sin(\phi)d\phi = -2\pi r^2 \cos(\phi)\right|_{0}^{\phi_0} = 2\pi r^2\left(1 - \cos(\phi_0)\right) $$

In this case, $\phi_0$ satisfies that $z = r\cos(\phi) = 2\cos(\phi_0) = 1 \rightarrow \cos(\phi_0) = \frac{1}{2}$ and thus:

$$ A = 2\pi 2^2\left(1 - \frac{1}{2}\right) = 4\pi $$

So I agree with your answer...

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