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 $x^2+y^2+z^2=1$
 A: 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?
A: Actually, your answer is almost correct.
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).
$$\iiint_D (x^2+y^2+z^2) \,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z = \int_{0}^{1} \int_{0}^{\pi} \int_{0}^{2\pi} r^4\sin\theta \,\mathrm{d}\phi \,\mathrm{d}\theta \,\mathrm{d}r \\= \int_{0}^{1} r^4 \,\mathrm{d}r\int_{0}^{\pi} \sin\theta\,\mathrm{d}\theta\int_{0}^{2\pi}  \,\mathrm{d}\phi \\= \frac{1}{5} \cdot 2 \cdot 2\pi=\frac{4\pi}{5}$$
