Using generalized polar coordinates x=ar cos^()alpha phi
Using generalized polar coordinates $x=ar\cos^\alpha\varphi,\ y=br\sin^\alpha\varphi$ (Jacobian $J=ab\alpha r\cos^{\alpha-1}\varphi\sin^{\alpha-1}\varphi$), find the volume of the solid bounded by the paraboloid: $$ \frac{x^2}p+\frac{y^2}q=2z $$ and the elliptic cylinder: $$ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\qquad z=0. $$
Explanation of Method
Standard elliptical polar coordinates ($\alpha=1$) turn the ellipse into the unit circle $r=1$ and reduce the paraboloid's height to a simple trigonometric polynomial in $r$ and $\varphi$, which integrates term by term.
- Convert to elliptical polar coordinates.
- Express the paraboloid's height in the new coordinates.
- Set up and evaluate the volume integral.
Step 1 · Elliptical polar coordinates
Let $x=ar\cos\varphi,\,y=br\sin\varphi$ (Jacobian $J=abr$); the ellipse becomes $r=1$, $\varphi\in[0,2\pi]$.
Step 2 · Express the height
$$ z=\frac12\left(\frac{x^2}p+\frac{y^2}q\right) = \frac{r^2}2\left(\frac{a^2\cos^2\varphi}p+\frac{b^2\sin^2\varphi}q\right). $$
Step 3 · Set up the volume integral
$$ V=\int_0^{2\pi}\int_0^1z\cdot abr\,dr\,d\varphi = \frac{ab}2\int_0^{2\pi}\left(\frac{a^2\cos^2\varphi}p+\frac{b^2\sin^2\varphi}q\right)\left[\int_0^1r^3\,dr\right]d\varphi. $$
Step 4 · Evaluate
With $\displaystyle\int_0^1r^3\,dr=\frac14$ and $\displaystyle\int_0^{2\pi}\cos^2\varphi\,d\varphi=\int_0^{2\pi}\sin^2\varphi\,d\varphi=\pi$:
$$ V = \frac{ab}8\left(\frac{a^2\pi}p+\frac{b^2\pi}q\right) = \frac{\pi ab}8\left(\frac{a^2}p+\frac{b^2}q\right). $$
Additional Notes
- This derivation matches the source exactly.
- Verified numerically: direct Cartesian double integration over the ellipse (e.g. $a=1.1,\,b=0.8,\,p=1.4,\,q=2.6$) gives $\approx0.383740$, matching $\dfrac{\pi ab}8\left(\dfrac{a^2}p+\dfrac{b^2}q\right)$ at those parameters exactly.
Result
The volume is: