# basic triple integral

I am pretty confident I can solve this question so please don't give me the answer, but I am having trouble "imagining" the area they are referring to.

Question: calculate $$\iiint_D (x^2+y^2+z^2)\,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z$$ where $D$ is the area confined by the 4 surfaces:

$$\begin{cases} x=0 \\ y=0 \\ z=0 \\ x^2+y^2+z^2=1 \end{cases}$$

I don't understand how to visualize this area. Isn't that just the sphere? just, anything inside the sphere? I don't see how the $xy,xz,yz$ add / subtract anything from the area of the sphere. they could have just omitted those surfaces and just said $D$ is there sphere $x^2+y^2+z^2=1$

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$x^2+y^2+z^2=1$ is the unit sphere, the three other equations define planes. But it's quite ambiguous, there are 8 possibles such "confined" surfaces, each 1/8 of the sphere. –  Jean-Claude Arbaut May 7 at 6:25
Yes I agree, but when they say space confined between those planes and the sphere, doesn't it mean just the union of all those 1/8 of the sphere? which is - the entire sphere? –  Oria Gruber May 7 at 6:26
Use spherical coordinates. –  Martín-Blas Pérez Pinilla May 7 at 6:26
What I would do is 1/8 of the sphere, but as I said, it's ambiguous. It would be more accurate to write for example $x\geq0$, $y\geq0$, $z\geq0$ and $x^2+y^2+z^2\leq1$. Then it would be an intersection of volumes, which is a well defined volume. –  Jean-Claude Arbaut May 7 at 6:28
The way it is written is ambiguous. The three first equations are planes, whereas the last equation is the (unit) sphere. Was there any indication in the question as to what octant you were working in? Generally it is advised to write something like $x\geq{0}$ to notify the individual as to where the region lives. Currently the region is the unit sphere cut up into 8 regions, but no one region is identified. –  DrkVenom May 7 at 15:30

First, notice $x^2+y^2+z^2=r^2$, this prevents you from simplifying the sum in spherical coordinates (but it's not difficult).

$$\iiint_D (x^2+y^2+z^2) \,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z = \int_{0}^{1} \int_{0}^{\pi} \int_{0}^{2\pi} r^4\sin\theta \,\mathrm{d}\phi \,\mathrm{d}\theta \,\mathrm{d}r \\= \int_{0}^{1} r^4 \,\mathrm{d}r\int_{0}^{\pi} \sin\theta\,\mathrm{d}\theta\int_{0}^{2\pi} \,\mathrm{d}\phi \\= \frac{1}{5} \cdot 2 \cdot 2\pi=\frac{4\pi}{5}$$

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Assuming the meaning is to the entirety of the space confined inside the unit sphere, we will move to spherical coordinates:

$x=r\sin\theta \cos\phi$, $y=r\sin\theta \sin \phi$, $z=r\cos\theta$

$\iiint_D (x^2+y^2+z^2) \,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z \\= \int_{0}^{1} \int_{0}^{\pi} \int_{0}^{2\pi}(r^2\sin^2\theta \cos^2\phi+r^2\sin^2\theta \sin^2 \phi+r^2\cos^2\theta)r^2\sin\theta \,\mathrm{d}\phi \,\mathrm{d}\theta \,\mathrm{d}r \\= \int_{0}^{1} \int_{0}^{\pi} \int_{0}^{2\pi}r^4\sin\theta \,\mathrm{d}\phi \,\mathrm{d}\theta dr = \frac{4\pi}{3}$

where $r^2\sin\theta$ is the jacobian.

Is this correct?

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If you integrate on the whole sphere, I don't quite understand why there are planes in the question... But the good news is, once you have the integral on the whole sphere, by symmetry the integral on 1/8th of the sphere is simply $\pi/10$, or 1/8th of your integral above - with correction of a little mistake, it's a 5 in denominator, not a 3, because the integral of $r^4$ is $r^5/5$. –  Jean-Claude Arbaut May 7 at 9:08