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Let B be the region in $R^3$ described by the inequalities $o\le x, 0\le y, 0\le z$ and $x^2+y^2+z^2\le 4$, then calculate the triple integral: $\int_B \sqrt{x^2+y^2+z^2}dxdydz$

Here's what I have tried: I've decided to split the integral to represent each variable.

$\int_0^2\int_0^{\sqrt{4-x^2}}\int_0^{\sqrt{4-y^2-x^2}}\sqrt{x^2+y^2+z^2}dzdydx$

Now converting this into spherical coordinates, I find that $0\le \theta\le \pi/2$ when $0\le x \le 2$ which is in quadrant 1.

However, I'm unsure on how to proceed with the calculating of $\phi$ and $r$ so I'd greatly appreciate the communities support on the next approach.

Stackcans
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2 Answers2

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In spherical coordinates, your integral becomes\begin{align}\int_0^{\pi/2}\int_0^{\pi/2}\int_0^2\rho^3\sin(\varphi)\,\mathrm d\rho\,\mathrm d\varphi\,\mathrm d\theta&=\frac\pi2\left(\int_0^{\pi/2}\sin(\varphi)\,\mathrm d\varphi\right)\left(\int_0^2\rho^3\,\mathrm d\rho\right)\\&=2\pi.\end{align}

José Carlos Santos
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We have $x \geq 0, y \geq 0, z \geq 0$ and so we are in the first octant.

If you are using $~ x = \rho \cos \theta \sin \phi, y = \rho \sin\theta \sin\phi, z = \rho \cos\phi$

$x^2 + y^2 + z^2 = \rho^2 \leq 4 \implies \rho \leq 2$

Your limits for $\theta$ is correct given we are in first octant.

As we know limits of $\phi$ is $0$ to $\pi$ measured against the positive z-axis. So for $z \geq 0, 0 \leq \phi \leq \pi/2$ and for $z \leq 0, \pi/2 \leq \phi \leq \pi$. You can see it clearly from $z = \rho \cos\phi$. $\cos\phi$ is positive for $0 \lt \phi \lt \frac{\pi}{2}$ and negative for $\frac{\pi}{2} \lt \phi \lt \pi$.

Math Lover
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  • This careful explanation is really helpful! I was slightly thrown off with $\phi$ and $\rho$ after looking at a few examples. Though this has cemented how I should be approaching $\rho$, however some doubt still remains on $\phi$. I seem to miss how you infer the boundaries from $z = \rho \cos \phi$, and why $\phi$ is from 0 to $\pi$. I'm looking at a schematic to better help me understand this on the direction that $\phi$ takes though I'm still having a few difficulties getting the picture. – Stackcans Oct 03 '21 at 21:05
  • @Stackcans sure I will explain. Take a ray from the origin (center of the sphere). When the ray is in XY-plane, what angle does it make with the positive z-axis? – Math Lover Oct 03 '21 at 21:08
  • @Stackcans See this for example https://www.geogebra.org/3d/xfwksx8t. This is a ray that makes $45^\circ$ with the positive $z$-axis. Now if the ray is in XY-plane, it will make $90^\circ$ with the positive $z$-axis, But when the ray goes below XY-plane (negative $z$), the angle it makes with the positive z-axis is $\gt 90^\circ$. Is it clear so far? – Math Lover Oct 03 '21 at 21:14
  • This explanation is very new to me however I'm thoroughly enjoying the learning experience. I sort of understand, is it at $90^o$ in the XY plane because of it representing an upper limit? whereas, when we go negative, this reverses the boundary limits so that $\pi/2<\phi<\pi$, would this be because the ray is further away from the z-axis when it's at a negative angle hence requiring a greater angle when the z-axis is positive? – Stackcans Oct 03 '21 at 21:24
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    Yes that is correct. If you take any line segment from the center (origin) in XY-plane, it makes $90^\circ$ with positive z-axis. By definition of spherical coordinates, $0 \leq \phi \leq \pi$ as the polar angle ($\phi$) is measured with positive z-axis and $0 \leq \theta \leq 2\pi$ as azimuthal angle $(\theta)$ is measured in XY plane or parallel to it. – Math Lover Oct 03 '21 at 21:28
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    Think of a ray from the origin making $\phi$ angle to the positive z-axis and the ray is above x-axis ($\theta = 0$). Now you spin it around the z-axis by $360^\circ$ counter-clockwise. Now if you take all such rays between $0 \leq \phi \leq \pi$ and rotate by $2 \pi$, you cover the whole sphere. You may be able to now see why limits of $\phi$ between $0$ and $\pi$ works. – Math Lover Oct 03 '21 at 21:40
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    I need to go to bed now but you think thru this further and we can get into a chat tomorrow to resolve rest of the questions. It does take time to sink in but if you try visualizing it, you will eventually get it. When I went schooling, it was never explained how spherical coordinates really works - the visualization part. I figured out gradually over time. – Math Lover Oct 03 '21 at 21:47
  • I thoroughly enjoyed our discussion and it will take some time as you have pointed out for me to grasp the picture fully. However, I feel far more mature towards understanding spherical coordinates in our conversation! I understand - I'm currently learning as a statistician so I have no mathematics background at school besides from self-studying day-in day-out non-stop and have found that only through practicing consistently, at some point I would be re-introduced with an old concept I once battled hard with only to find it to be the most simple! :) – Stackcans Oct 03 '21 at 21:51