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I am trying to solve the following question:

The volume of the solid $E$ can be represented as $$\int_{-3}^3 \int_0^{\sqrt{9-x^2}} 3 - \sqrt{x^2+y^2} dydx$$ Describe the solid in spherical coordinates.

I graphed the solid and it looks like this:

So clearly $0 \le \theta \le \pi$ and $0 \le \phi \le \frac{\pi}{2}.$ Also, for each $(\theta, \phi)$ the value of $\rho$ ranges from $0$ until the surface of the cone. The cone is $z = 3-\sqrt{x^2+y^2}$. Using the substitutions \begin{align*} x &= \rho \cos \theta \sin \phi\\ y &= \rho \sin \theta \sin \phi\\ z &= \rho \cos \phi \end{align*} we get \begin{align*} \rho \cos \phi &= 3 - \sqrt{\rho^2 \sin^2 \phi}\\ \rho &= \frac {3}{\cos \phi + \sin \phi} \end{align*} So my answer is $$E_{\text{spherical}} = \{(\rho, \theta, \phi)| 0 \le \theta \le \pi, 0 \le \phi \le \frac{\pi}{2}, 0 \le \rho \le \frac {3}{\cos \phi + \sin \phi} \}$$ However, the correct answer is $$E_{\text{spherical}} = \{(\rho, \theta, \phi)| 0 \le \theta \le \pi, 0 \le \phi \le \frac{\pi}{4}, 0 \le \rho \le 3 \csc\phi \}$$

Did I make any mistake?

Thanks!

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2 Answers 2

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The solid is half of an inverted cone with vertex at $(0, 0, 3)$ and above $z = 0$.

Equation of the surface of the cone is $ ~\sqrt{x^2+y^2} = 3 - z$

That translates to $ \rho \sin\phi = 3 - \rho \cos\phi \implies \rho = \frac{3}{\cos\phi + \sin\phi}$

Please note that $0 \leq \phi \leq \pi/2$ as we are above $z = 0$.

Also given bounds of $x$ and $y$, you can see that the projection of the solid in xy-plane is in the first and the second quadrant. That leads to $0 \leq \theta \leq \pi$.

$E_{\text{spherical}} = \{(\rho, \theta, \phi)| 0 \le \rho \le \frac{3}{ \cos\phi + \sin\phi}, 0 \le \theta \le \pi, 0 \le \phi \le \frac{\pi}{2} \}$

But to evaluate the volume, it is easier to use $ ~ x = \rho \cos\theta\sin\phi, y = \rho \sin\theta \sin\phi, z = 3 - \rho \cos\phi$

That translates the cone to $ ~ \phi = \frac{\pi}{4}$

$\rho$ is bound by the plane $z = 3 - \rho \cos\phi = 0 \implies \rho = 3 \sec\phi$

So the integral to find volume in spherical coordinates should be,

$ \displaystyle \int_0^{\pi} \int_0^{\pi/4} \int_0^{3\sec\phi} ~ \rho^2 \sin\phi ~ d\rho ~ d\phi ~ d\theta$

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  • $\begingroup$ The OP did not ask about the volume or the integral that would express it... It was about describing the region in spherical coordinates. $\endgroup$ Nov 9, 2021 at 15:39
  • $\begingroup$ Oh wow this makes a lot of sense, thanks! $\endgroup$
    – user56202
    Nov 9, 2021 at 15:41
  • $\begingroup$ @PierreCarre From the bounds on the integral I can see the region. $\endgroup$
    – user56202
    Nov 9, 2021 at 15:42
  • $\begingroup$ @PierreCarre sure I would add that too $\endgroup$
    – Math Lover
    Nov 9, 2021 at 15:43
  • $\begingroup$ @user56202 You could see the region right from the start! The bounds do not correspond to the region, although it gives you the correct value for the volume. $\endgroup$ Nov 9, 2021 at 15:43
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The "correct" answer does not seem to be correct, at least not according to the change of variable you present ($\phi = 0$ for points in the $z$ axis). It should be:

$$ \{(\theta, \phi, \rho): 0\leq \theta \leq \pi, 0 \leq \phi \leq \frac{\pi}{2}, 0 \leq \rho \leq \frac{3}{\cos \phi+ \sin \phi} \}, $$

as you mention. So, your solution is correct.

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  • $\begingroup$ Thanks. But why is $\phi$ bounded below by $\pi/4$? Our solid contains points on the $z$ axis, which have a $\phi$ of $0$. $\endgroup$
    – user56202
    Nov 9, 2021 at 15:33
  • $\begingroup$ @user56202 My bad! I'll correct! $\endgroup$ Nov 9, 2021 at 15:34

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