How to find the volume by using shell method

Question

I would like to find the volume of $\frac {x^2} 9+y^2=1$ rotated around the $x$ axis, but using the shell method. How can I do this?

Letting the thickness be $dy$ and the problem I encountered is finding the radius.

Answer 1

Disc method: Your ellipse gives $y=\pm \sqrt{1-\frac{x^2}9}$

Since your are rotating about the $x$-axis, this gives a radius of $\sqrt{1-\frac{x^2}9}$, a disc circular area of $\pi r^2$ and so a volume of $$\int_{-3}^3 \pi\left(1-\frac{x^2}9\right) dx $$

---

Shell method: Your ellipse gives $x=\pm 3\sqrt{1-y^2}$

Since your are rotating about the $x$-axis, this gives a cylindrical "height" of $6\sqrt{1-y^2}$ and radius of $y$, a cylindrical shell surface area of $2\pi r h$ and so a volume of $$\int_{0}^1 12\pi y\sqrt{1-y^2}\, dy $$

---

If you do the integrations you get the same answer. You would also get the same answer taking a sphere of radius $1$ and then stretching it by a factor of $3$ along the $x$-axis to give your ellipsoid