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This is the question:

The solid generated by rotating the region inside the ellipse with equation $$ \left( \frac{x}{a} \right)^2 + \left( \frac{y}{b} \right)^2 = 1 $$ around the $x$-axis is called an ellipsoid.

(a) Show that the ellipsoid has volume $\displaystyle \frac{4}{3} \pi a b^2.$

(b) What is the volume if the ellipse is rotated around the $y$-axis.

(I want to accomplish this using integrals and basic plane geometry. To give you an idea of how much I know about integrals I'm $4$ weeks into my calculus II course, which is my first exposure to integrals)

I started by drawing an ellipse on a Cartesian plane. The ellipse went from $a$ to $b$, I then rotated this ellipse around the $x$-axis to get an ellipsoid. Now to get the volume I have to find a cross-sectional area, so I noticed that the ellipsoid is really made up of a bunch of circles stacked along the $x$-axis. Where I'm stuck at right now is how I can find the radius of these circles, which would give me my integrand.

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  • $\begingroup$ @Downvoters, how do you suggest that I improve my question? $\endgroup$ – Dunka Sep 25 '14 at 22:07
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I'll use disk integration, where the volume is given by; $$\pi \int_{x_1}^{x_2} f(x)^2 \mathrm{d}x$$

First we solve for $y$ as a function of $x$, which is trivial so I'll just give the result; $$y = b\sqrt{1-\left (\frac{x}{a}\right )^2}$$

Now the interesting part. The ellipse doesn't go from $a$ to $b$ but from $-a$ to $a$. So we construct;

$$\pi \int_{-a}^{a} b^2 \left (1-\left(\frac{x}{a}\right )^2 \right) \mathrm{d}x=\pi \left( -\frac{b^2}{a^2} \int_{-a}^{a} x^2 \mathrm{d}x + b^2 \int_{-a}^{a} \mathrm{d}x \right )=\pi\left (-\frac{2ab^2}{3}+2ab^2\right )=\frac43 \pi a b^2$$

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  • $\begingroup$ Thank you so much. Why does the ellipse go from -a to a though? $\endgroup$ – Dunka Sep 25 '14 at 22:28
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    $\begingroup$ Because it's horizontal intercepts are at $\pm a$. $\endgroup$ – UserX Sep 25 '14 at 22:41
  • $\begingroup$ And how did you figure that out? $\endgroup$ – Dunka Sep 25 '14 at 22:44
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    $\begingroup$ That's textbook knowledge. You could prove it yourself, by setting $y=0$ but it's faster to learn some fixed points. $\endgroup$ – UserX Sep 25 '14 at 22:46
  • $\begingroup$ Would the area be the same if I rotated the ellipse around the y-axis? I did some work and it said it would be. Wondering if I'm right or not? $\endgroup$ – Dunka Sep 25 '14 at 23:37
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The radius of those circles is the height of the function. Here, the radius is $$y = b\sqrt{1-(\frac{x}{a})^2}.$$

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We know that the volume of the solid generated by rotating the curve $y=f(x)\space $ about the x-axis is given as follows $$V=\int_{x_1}^{x_2}\pi y^2dx$$ Hence, the volume of the solid generated by rotating the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \space$ about the x-axis is obtained by setting $y^2=\frac{b^2}{a^2} (a^2-x^2) \space$ in the above formula & applying the proper limits $$V=\pi \int_{-a}^{a} \frac{b^2}{a^2} (a^2-x^2)dx= \frac{\pi b^2}{a^2} \int_{-a}^{a} (a^2-x^2)dx$$ $$=\frac{2\pi b^2}{a^2} \int_{0}^{a} (a^2-x^2)dx=\frac{2\pi b^2}{a^2} \left[a^2x-\frac{x^3}{3} \right]_{0}^{a}$$ $$=\frac{2\pi b^2}{a^2} \left[a^3-\frac{a^3}{3} \right]=\frac{2\pi b^2}{a^2} \left[\frac{2a^3}{3} \right]=\frac {4}{3}\pi ab^2$$ We know that the volume of the solid generated by rotating the curve $y=f(x)\space $ about the y-axis is given as follows $$V=\int_{y_1}^{y_2}\pi x^2dy$$

Hence, the volume of the solid generated by rotating the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \space$ about the y-axis is obtained by setting $x^2=\frac{a^2}{b^2} (b^2-y^2) \space$ in the above formula & applying the proper limits $$V=\pi \int_{-b}^{b} \frac{a^2}{b^2} (b^2-y^2)dy= \frac{\pi a^2}{b^2} \int_{-b}^{b} (b^2-y^2)dy$$ $$=\frac{2\pi a^2}{b^2} \int_{0}^{b} (b^2-y^2)dy=\frac{2\pi a^2}{b^2} \left[b^2y-\frac{y^3}{3} \right]_{0}^{b}$$ $$=\frac{2\pi a^2}{b^2} \left[b^3-\frac{b^3}{3} \right]=\frac{2\pi a^2}{b^2} \left[\frac{2b^3}{3} \right]=\frac {4}{3}\pi a^2 b$$

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