# Integration with substitution to cylindrical coordinate

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

• Where is the variable $h$ coming from? Also if you put a backslash before $sin$ and $cos$, they turn into $\sin{t}$ and $\cos{t}$. – Calvin Godfrey Jan 7 at 17:03
• @CalvinGodfrey $h$ equals $z$, I've fixed it. – J. Doe Jan 7 at 17:04
• Change first to $u=x/a$, $v=y/b$, $w=z/c$ and then to spherical coordinates. – A.Γ. Jan 7 at 17:06

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.