I'm having trouble setting up the bounds and the integral. The question asked to find the mass of the 3D region $ x^2 +y^2 + z^2 ≤ 4,x ≥ 0, y≥ 0, z ≥ 0 $ if the density is equal to xyz. I don't want to evaluate it, I just want to clear some few things so I may never have trouble with such problems again.
Can this question be done with Polar coordinates? If yes, then why and what will be the limits?
In case of rectangular coordinates, how would the limits be set up and what would be the integrand?
My attempt:
To find x and y intercepts, taking y and z = 0, so we have $x = 2$, $-2$. Are these the limits of x? I don't think that -2 will be the lower limit because the region specifies that x ≥ 0 so the limits will be [0,2]. Kindly confirm.
Using the same logic for y and making y depend on x, the limits will be $ y = [0,\sqrt{4-x^2}] $ disregarding the out of bound limit $-\sqrt{4-x^2} $ . Is this a correct assumption?
Finally the limits for z will be $[0, \sqrt{4-x^2-y^2}]$
Now I'm not sure how to set up the integral but I would assume that it would be in the order of dzdydx with the integrand being 1 but coming back to the question, I have integrate xyz over the region to calculate the mass. Thank you in advance.
Since $x=r\sin\theta\cos\varphi,\,y=r\sin\theta\sin\varphi,\,z=r\cos\theta$ has Jacobian $r^2\sin \theta$, an infinitesimal region has mass $xyz dxdydz=r^5 dr\sin^3\theta\cos\theta d\theta\sin\varphi\cos\varphi d\varphi$. We integrate $r$ from $0$ to $2$, $\theta$ from $0$ to $\pi/2$ and $\varphi$ from $0$ to $\pi/2$, obtaining (if my arithmetic's right) $$\int_0^2 r^5 dr\int_0^{\pi/2}\sin^3\theta\cos\theta d\theta\int_0^{\pi/2}\frac{1}{2}\sin 2\varphi d\varphi=\frac{2^6}{6}\frac{1}{4}\frac{1}{4}=\frac{2}{3}.$$