Volume of ellipsoid in Cartesian co ordinates ( w/o changing to spherical or cylindrical systems) I tried triple integrating over the $$\int_{-a}^{a} \int_{-b\sqrt{1-\frac{x^2}{a^2}}}^{b\sqrt{1-\frac{x^2}{a^2}}} \int_{-c\sqrt{1-\frac{x^2}{a^2}-\frac{y^2}{b^2}}}^{c\sqrt{1-\frac{x^2}{a^2}-\frac{y^2}{b^2}}} {d}z {d}y {d}x $$ but getting $4/3 π a^2 b c$. Checking the possibility of deriving it with Cartesian coordinates and not converting to spherical or cylindrical systems. The derivation is as below
$$ z = \pm c\sqrt{1-\frac{x^2}{a^2}-\frac{y^2}{b^2}} ; y = \pm b\sqrt{1-\frac{x^2}{a^2}} \{z=0\}; x = \pm \,\, a \{z=0,y=0\}$$
for simplicity let $z = \pm c_1 \, and \, y = \pm b_1$
$$ V_E = \int_{-a}^{a}\int_{-b_1}^{b_1} z\Big|_{-c_1}^{c_1}\,{d}y{d}x$$
evaluating for z
$$ V_E = \int_{-a}^{a}\int_{-b_1}^{b_1} c\sqrt{1-\frac{x^2}{a^2}-\frac{y^2}{b^2}}\,-\,\left[ -c\sqrt{1-\frac{x^2}{a^2}-\frac{y^2}{b^2}}\right] {d}y{d}x$$
$$ V_E = \int_{-a}^{a}\int_{-b_1}^{b_1} 2c\sqrt{1-\frac{x^2}{a^2}-\frac{y^2}{b^2}}\,{d}y{d}x$$
which is
$$V_E = \frac{2c}{ab} \int_{-a}^{a}\int_{-b_1}^{b_1} \sqrt{{a^2}{b^2}-{x^2}{b^2}-{a^2}{y^2}}\,{d}y{d}x$$
now we know that 
$$\boxed {\int \sqrt{{\alpha^2}-{\chi^2}}\,{dx} = \frac{x}{2}\sqrt{{\alpha^2}-{\chi^2}} +\frac{\alpha^2}{2}sin^{-1}\left(\frac{\chi}{\alpha}\right) + C\,\,}$$
putting $\alpha = \sqrt{{a^2}{b^2}-{x^2}{b^2}}\, and \, \chi = \sqrt{{a^2}{y^2}}$ in above, we get
$$V_E = \frac{2c}{ab} \int_{-a}^{a}\, \frac{ay}{2}\sqrt{(ab)^2-(bx)^2-(ay)^2}+\frac{(ab)^2-(bx)^2}{2}sin^{-1}\left(\frac{ay}{\sqrt{(ab)^2-(bx)^2}}\right)\Big|_{-b_1}^{b_1}\,$$
Substituting the limits;
$$V_E = \frac{2c}{ab} \int_{-a}^{a}\,\left[\frac{a}{2}\left(b\sqrt{1-\frac{x^2}{a^2}}\right)\right]\sqrt{(ab)^2-(bx)^2-a^2\left(b^2\left(1-\frac{x^2}{a^2}\right)\right)} + \frac{(ab)^2-(bx)^2}{2}\,sin^{-1} \left(\frac{a}{\sqrt{(ab)^2-(bx)^2}}b\sqrt{1-\frac{x^2}{a^2}}\right) - \,\left[\frac{a}{2}\left(-b\sqrt{1-\frac{x^2}{a^2}}\right)\right]\sqrt{(ab)^2-(bx)^2-a^2\left(b^2\left(1-\frac{x^2}{a^2}\right)\right)} - \frac{(ab)^2-(bx)^2}{2}\,sin^{-1} \left(\frac{a}{\sqrt{(ab)^2-(bx)^2}}(-b)\sqrt{1-\frac{x^2}{a^2}}\right) dx$$
Simplifying
$$V_E = \frac{2c}{ab} \int_{-a}^{a}\,\frac{b}{2}\,\sqrt{a^2-x^2}\,(0) + \frac{b^2(a^2-x^2)}{2}\,sin^{-1}(1)\,-\,\,\frac{-b}{2}\,\sqrt{a^2-x^2}\,(0) - \frac{b^2(a^2-x^2)}{2}\,sin^{-1}(-1) dx$$
we know that $sin^{-1}(1) = \frac{\pi}{2}; sin^{-1}(-1) = \frac{-\pi}{2} $
$$V_E = \frac{2c}{ab} \int_{-a}^{a}\,0 + \frac{b^2(a^2-x^2)}{2}\,\frac{\pi}{2}-\,0 - \frac{b^2(a^2-x^2)}{2}\,\left(\frac{-\pi}{2}\right)dx$$
taking the constants out;
$$V_E = \frac{2\pi b^2c}{2ab} \int_{-a}^{a}(a^2-x^2)\,dx$$
Solving for x
$$V_E = \frac{\pi bc}{a}\left( a^2x - \frac{x^3}{3} \Big|_{-a}^{a} \right)$$
$$V_E = \frac{\pi bc}{a}\left(\left[ a^3 - \frac{a^3}{3}\right]-\left[- a^3 - \frac{-a^3}{3}\right]\right)$$
hence
$$V_E = \frac{4\pi a^2bc}{3}$$
Not sure what went wrong
 A: For starters, I would rescale $x, y, z$ by $X = x/a, \ Y = y/b, \ Z = z/c$. Thus our triple integral becomes
$$ I =  abc \int_{-1}^1 \int_{-\sqrt{1 - X^2}}^{\sqrt{1 - X^2}} \int_{-\sqrt{1 - X^2 - Y^2}}^{\sqrt{1 - X^2 - Y^2}} dZ dY dX.  $$
Integrating with respect to $Z$
\begin{align*}
I & = 2abc \int_{-1}^1 \int_{-\sqrt{1 - X^2}}^{\sqrt{1 - X^2}} \sqrt{1 - X^2 - Y^2} \ dY dX \\
& = \pi a b c \int_{-1}^1 1- X^2 \ dX \\
& = 2 \pi a b c \left (X - \frac{X^3}{3} \right ) \bigg|_{X=0}^{X=1}  \\
& = \frac{4 \pi a b c}{3}. 
\end{align*}
A: Let $V$ denote the volume. You can simplify things from the start by taking advantage of symmetry:
$$\begin{align*}
V&=\int_{-a}^a\int_{-b\sqrt{1-\frac{x^2}{a^2}}}^{b\sqrt{1-\frac{x^2}{a^2}}}\int_{-c\sqrt{1-\frac{x^2}{a^2}-\frac{y^2}{b^2}}}^{c\sqrt{1-\frac{x^2}{a^2}-\frac{y^2}{b^2}}}\,\mathrm dz\,\mathrm dy\,\mathrm dx\\[1ex]
&=8\int_0^a\int_0^{b\sqrt{1-\frac{x^2}{a^2}}}\int_0^{c\sqrt{1-\frac{x^2}{a^2}-\frac{y^2}{b^2}}}\mathrm dz\,\mathrm dy\,\mathrm dx
\end{align*}$$
The integral with respect to $z$ is trivial:
$$V=8c\int_0^a\int_0^{b\sqrt{1-\frac{x^2}{a^2}}}\sqrt{1-\frac{x^2}{a^2}-\frac{y^2}{b^2}}\,\mathrm dy\,\mathrm dx$$
Substitute $y=b\sqrt{1-\frac{x^2}{a^2}}\sin\theta$, treating $x$ as a constant, so that $\mathrm dy=b\sqrt{1-\frac{x^2}{a^2}}\cos\theta\,\mathrm d\theta$. Under this substitution, we have
$$y=0\implies\sin\theta=0\implies\theta=0$$
$$y=\sqrt{1-\frac{x^2}{a^2}}\implies\sin\theta=1\implies\theta=\frac\pi2$$
$$\begin{align*}
V&=8c\int_0^a\int_0^{\frac\pi2}\sqrt{\left(1-\frac{x^2}{a^2}\right)-\frac1{b^2}\left(b\sqrt{1-\frac{x^2}{a^2}}\sin\theta\right)^2}\,b\sqrt{1-\frac{x^2}{a^2}}\cos\theta\,\mathrm d\theta\,\mathrm dx\\[1ex]
&=8bc\underbrace{\left(\int_0^a1-\frac{x^2}{a^2}\,\mathrm dx\right)}_{\frac{2a}3}\underbrace{\left(\int_0^{\frac\pi2}\cos^2\theta\,\mathrm d\theta\right)}_{\frac\pi2}\\[1ex]
&=\frac{4\pi abc}3
\end{align*}$$
Not sure where your mistake occurs, but you're obviously missing a factor of $a$ in the leading fractions of the integral leading up to your result. I suspect it's because of the way you were rewriting the square root terms, which I'd suggest not doing if you don't have to.
