Find the volume of revolution of an ellipse about y-axis.

Question

Find the volume of revolution of an ellipse about $y-$axis: $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ is the given equation. I tried to know that the volume of revolutions are same about $x$ and $y$ axis?

Answer 1

Volumes of ellipsoid of revolution about $y,x$ axes are respectively $$ \frac43 \pi a^2 b ,\quad \frac43 \pi a b^2\,$$

During integration in first case $x$ varies between limits $(0,a)$, in second case $y$ varies between limits $(0,b)$. This procedure gives an expectation basically for volume results to be different.

From a physical understanding also a narrow spindle $ a>>b$ has less volume when swept around $x$ axis compared to $y$ axis.

For them to be equal, it should be $ \,a=b, $ i.e., when it is the case of sphere with equal axes.

Answer 2

The volume of revolution around the $y$-axis for a function $y(x)$ running from $x_1$ to $x_2$ is given by $$\pi\int_{x_1}^{x_2}y^2 \mathrm{d}x.$$

Now, if we simple solve for $y$ in your equation, we don't get a proper function, but rather an implicit one. However, the desired volume will simply be two times the volume of the rotated upper half of the ellipse, and this upper half part, we can define a function, $y(x)$, namely $$y(x)=+b\sqrt{1-\frac{x^2}{a^2}}.$$

The limits $x_1,x_2$ of your ellipse should now be found in terms $a$ and $b$.

Have a go at solving the rest of the problem, and let me know if you're still stuck after that. Good luck!