# Finding the volume of a solid of revolution

Could someone help me with this?

Find the volume of the solid generated by revolving about the y-axis the region bounded by the upper half of the ellipse $$x^2/a^2 + y^2/b^2 = 1$$ and the x-axis, and thus find the volume of a prolate spheroid.

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Since you are new: (i) Titles should be informative, not pleas for help; (ii) Many of us consider it very rude for people to post in the imperative (giving orders, assigning problems), as if you were the professor, we the students, and you are giving us homework to do. If you must quote, then please place it in a quote box. But in any case, you should always include (iii) the context (what course is this for? is this homework? Is this self-study? what? If it is homework, please add the [homework] tag); (iv) Say what you have managed to do, or where you are stuck and why; (cont) –  Arturo Magidin Jul 30 '11 at 3:55
(cont) (v) Be sure the question is complete and intelligible. In this case, you seem to have cut-n-pasted, and information was lost (note the last sentence is nonsensical). –  Arturo Magidin Jul 30 '11 at 3:55
"$a$ and $b$ are positive constants, with $a > b$ "??? My magic 8-ball might be off, though... –  The Chaz 2.0 Jul 30 '11 at 4:23
Since you are new here, you may want to read our FAQ on homework-type questions. –  Willie Wong Jul 30 '11 at 4:26
Hint: The volume is $\int_{-a}^a \pi y^2\;dx$. –  André Nicolas Jul 30 '11 at 4:58

We use the Method of Slicing. Let $f(x) \ge 0$ on the interval $[a,b]$, and let $R$ be the region below the curve $y=f(x)$, above the $x$-axis, from $x=p$ to $x=q$. Then the volume of the solid obtained by rotating $R$ about the $x$-axis is $$\int_p^q \pi(f(x))^2\;dx.$$

In our case, $R$ is the region below the top half of our ellipse.

For that ellipse, we have $$\frac{y^2}{b^2}=1-\frac{x^2}{a^2}$$ and therefore $$f(x)=b\sqrt{1-\frac{x^2}{a^2}}.$$ The top half of the ellipse meets the $x$ axis at $x=\pm a$. So our area is $$\int_{-a}^a\pi\left(b\sqrt{1-\frac{x^2}{a^2}}\right)^2\;dx.$$

We can simplify the calculation by noting that the ellipse is symmetrical about the $y$-axis. So we integrate from $x=0$ to $x=a$, and double the result. Thus we want $$2\int_0^a \pi \left(b^2-\frac{b^2x^2}{a^2}\right)\;dx.$$

The integration is easy. After a little while we get $$\frac{4}{3}\pi ab^2.$$

Note that if $b=a$, we get the familiar formula for the volume of a ball of radius $a$. This gives a useful partial check on the correctness of the calculation.

Intuition for the Volume Formula: Think of our solid as a somewhat peculiarly shaped salami, fat in the middle. Take a thin slice of the salami, perpendicular to the $x$-axis. Let the slice have thickness "$dx$", and suppose the slice was obtained by slicing through the salami at $x$ and at $x+dx$. Then the slice is almost perfectly round, and has radius roughly equal to $f(x)$. So the volume of the slice is roughly $\pi(f(x))^2\,dx$. Now let integration "add up" the volumes of these slices for us, as $x$ travels from $-a$ to $a$.

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