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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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put on hold as off-topic by Jonas Meyer, Chappers, graydad, Jeel Shah, Peter Woolfitt 2 days ago

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"$a$ 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

1 Answer 1

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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