Find the volume of the solid bounded by the elliptic
Find the volume of the solid bounded by the elliptic cylinder: $$ \left(\frac{x-a}a\right)^2+\left(\frac yb\right)^2=1 $$ and the parabolic cylinder: $$ z^2=x \qquad(a,b>0). $$
Explanation of Method
The ellipse is centered at $(a,0)$ with semi-axes $a$ (in $x$) and $b$ (in $y$), so $x$ ranges over $[0,2a]$. The parabolic cylinder gives $z=\pm\sqrt x$, so at each point of the elliptical base the solid has full height $2\sqrt x$. Substituting $x=a(1+u)$ turns the base into a clean interval in $u\in[-1,1]$.
- Set up the volume as $\displaystyle\iint_{\text{ellipse}}2\sqrt x\,dy\,dx$.
- Substitute $x=a(1+u)$ and integrate out $y$.
- Evaluate the resulting single integral in $u$.
Step 1 · Set up the volume integral
At fixed $x\in[0,2a]$, the ellipse gives $y\in\left[-b\sqrt{1-\left(\frac{x-a}a\right)^2},\,b\sqrt{1-\left(\frac{x-a}a\right)^2}\right]$, and $z$ ranges over $[-\sqrt x,\sqrt x]$. So:
$$ V=\iint_{\text{ellipse}}2\sqrt x\,dA = \int_0^{2a}2\sqrt x\cdot2b\sqrt{1-\left(\frac{x-a}a\right)^2}\,dx. $$
Step 2 · Substitute $x=a(1+u)$
With $x=a(1+u)$, $dx=a\,du$, and $\dfrac{x-a}a=u$, as $x$ runs over $[0,2a]$, $u$ runs over $[-1,1]$:
$$ V = \int_{-1}^1 4b\sqrt{a(1+u)}\sqrt{1-u^2}\cdot a\,du = 4ab\sqrt a\int_{-1}^1\sqrt{1+u}\sqrt{1-u^2}\,du = 4ab\sqrt a\int_{-1}^1(1+u)\sqrt{1-u}\,du, $$
using $\sqrt{1-u^2}=\sqrt{1-u}\sqrt{1+u}$.
Step 3 · Evaluate the remaining integral
Let $t=1-u$, so $u=1-t,\,du=-dt$, $t:2\to0$ as $u:-1\to1$:
$$ \int_{-1}^1(1+u)\sqrt{1-u}\,du = \int_0^2(2-t)\sqrt t\,dt = \left[\frac{4t^{3/2}}3-\frac{2t^{5/2}}5\right]_0^2 = \frac{4\cdot2\sqrt2}3-\frac{2\cdot4\sqrt2}5 = \frac{8\sqrt2}3-\frac{8\sqrt2}5 = \frac{16\sqrt2}{15}. $$
Therefore:
$$ V = 4ab\sqrt a\cdot\frac{16\sqrt2}{15} = \frac{64ab\sqrt{2a}}{15}. $$
Additional Notes
- This derivation matches the source exactly.
- Verified both symbolically (the intermediate integral $\int_{-1}^1(1+u)\sqrt{1-u}\,du=\frac{16\sqrt2}{15}$ evaluates exactly) and via a fully independent direct 2D Cartesian double integral (e.g. with $a=1.3,\,b=0.9$, direct numerical integration gives $\approx8.049358$, matching $\frac{64ab\sqrt{2a}}{15}$ evaluated at the same parameters).
Result
The volume is: