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Write a triple integral in sphereical coordinates for the volumne inside the cone $z^2=x^2+y^2$ and between the plane $z=1$ and $z=2$

$$x=r\sin(\theta) \cos(\Phi)$$

$$y=r\sin(\theta) \sin(\Phi)$$

$$z=r\cos(\theta)$$

$$0\le \Phi \le 2\pi$$

$$\frac 1 {\cos(\theta)} \le r \le \frac 2 {\cos(\theta)}$$

From searching online already I found that $0 \le \theta \le \pi/4$, but I have no idea why. Why wouldn't it be $\pi$ instead of $\pi /4$?

I found this link that describes the cone perfectly. But in this case, wouldn't it between $0 \le \theta \le \pi/2$?

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  • $\begingroup$ Well, this sphere is symmetrical. You only need to integrate to the factor of pi/4 $\endgroup$ – wesssg Dec 8 '16 at 23:01
  • $\begingroup$ Possible typo but I believe the upper bound for $r$ is $2/cos\theta$ $\endgroup$ – WaveX Dec 8 '16 at 23:18
  • $\begingroup$ Yes it is! Thank you for pointing it out! I have it written down correctly, just made an error while typing. $\endgroup$ – ChillingK Dec 8 '16 at 23:29
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Unfortunately, there are different ways of defining spherical coordinates, so I'll approach this carefully.

Judging from your bounds on $\Phi$, I assume you're using $\Phi$ to denote the angle swept out in the $xy$-plane. So that would mean you're using $\theta$ to denote the angle down from the positive $z$-axis. Is that all correct?

If so, $\theta=\pi$ would put you on the negative $z$-axis, and $0\le \theta\le\pi$ covers all of $3$-dimensional space. $0\le\theta\le\pi/2$ describes all of $3$-dimensional space where $z\ge0$, since $\theta=\pi/2$ puts you in the $xy$-plane. To visualize the region described by $0\le\theta\le\pi/4$, imagine a ray starting at the origin and pointing along the positive $z$-axis. Now rotate that ray $45^\circ$ toward the $xy$-plane, and rotate the result around the $z$-axis. That also happens to describe perfectly the cone $z^2=x^2+y^2$, so you're all set.

See the Wikipedia page on spherical coordinates for more detail.

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