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Solve the integral $$ \\A:= \int_{{x^2\over a^2}+{y^2\over b^2}+{z^2\over c^2}\leq 1}{x^2\over a^2}+{y^2\over b^2}+{z^2\over c^2}dxdydz \ $$


In my solution I have substituted to cylindrical coordinates ($x=r\operatorname{\cos} t,y=r\operatorname{sin} t,z=h$) with the ranges $-1\leq h\leq 1$ and $0\leq t\leq 2\pi$. And when I get fixed $h$ and $t$, then $r$ "moves" from $0$ until it meets a new elipse which determinated by $a'={{h\over c}\cdot a}, b'={{h\over c}\cdot b}$. So I need to find the meeting point of $r$ with the new elipse $1={x^2\over a'^2}+{y^2\over b'^2}=({r \cos t\over ah})^2+({r\sin t\over bh})^2$. Hence, if I sign $\lambda:=+{abh\over \sqrt{a^2\sin^2 t+b^2\cos^2 t}}$, then the required range of $r$ is from $0$ to $\lambda$. Hence, ($J=r$), $$ \\ A=\int_{-1}^1\int_0^{2\pi}\int_0^\lambda r({r^2cos^2 t\over a^2}\ +{r^2\sin^2 t\over b^2}+{h^2\over c^2})drdtdh \\ =...=\int_{-1}^1\int_0^{2\pi}(\lambda^4({\cos^2 t\over 4a^2}+{\sin^2 t\over 4 b^2})+\lambda^2{h^2\over c^2})dtdh\ \\=...=\int_{-1}^1\int_0^{2\pi}({a^6b^6c^2h^8\over 4(a^2\sin^2 t+b^2 \cos^2 t)^3}+{a^4b^4h^6\over c^2(a^2\sin^2 t+b^2\cos^2 t)^2})dtdh $$ Is this integral doable or I didn't do this exercise right? Thianks

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    $\begingroup$ Where is the variable $h$ coming from? Also if you put a backslash before $sin$ and $cos$, they turn into $\sin{t}$ and $\cos{t}$. $\endgroup$ – Calvin Godfrey Jan 7 at 17:03
  • $\begingroup$ @CalvinGodfrey $h$ equals $z$, I've fixed it. $\endgroup$ – J. Doe Jan 7 at 17:04
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    $\begingroup$ Change first to $u=x/a$, $v=y/b$, $w=z/c$ and then to spherical coordinates. $\endgroup$ – A.Γ. Jan 7 at 17:06
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Hint for a simpler way. Let $X=x/a$, $Y=y/b$ and $Z=z/c$, then $$\begin{align}\int_{{x^2\over a^2}+{y^2\over b^2}+{z^2\over c^2}\leq 1}&\left({x^2\over a^2}+{y^2\over b^2}+{z^2\over c^2}\right)dxdydz\\ &= abc\int_{{X^2}+{Y^2}+{Z^2}\leq 1}\left({X^2}+{Y^2}+{Z^2}\right)dXdYdZ\\ &=abc\int_{r=0}^1r^2 (4\pi r^2) dr \end{align}$$ where in the last step we used the spherical coordinates.

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