Find the volume $V$ of the solid generated by revolving the region bounded by the graph of, $\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1, a > 0, b > 0$, in the first quadrant, about the $x$-axis.

Solution The equation $\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1$ defines an ellipse. The intercepts of its graph are $(a,0), (0,b), (-a,0)$, and $(0,-b)$. The region to be revolved is the shaded region in the first quadrant shown in Figure 36(a).

The shell method.

In the shell method, when the region is revolved about the $x$-axis, we partition the $y$-axis and use horizontal shells. A revolution about the $x$-axis requires integration with

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respect to $y$, so we express the equation of the ellipse as $$x=g(y)=\dfrac{a}{b}\sqrt{~b^{2}-y^{2}}$$

The volume of a typical shell is $$V_{i}=2\pi\ \hbox{(Average radius)(Height)(Thickness)} = 2\pi v_{i}g(v_{i})\Delta y$$

See Figure 36(b).

Figure 36(c) shows the solid of revolution. The volume $V$ of the solid of revolution is $$\begin{eqnarray*} V &=&2\pi \int_{0}^{b}y~g(y)~{\it dy}=2\pi \int_{0}^{b}y\left( \dfrac{a}{b}\sqrt{ b^{2}-y^{2}}\right) {\it dy} \underset{\underset{\underset{\color{#0066A7}{\hbox{then \({du}={-2y}{ dy}\)}}}{\color{#0066A7}{\hbox{Let \(u=b^2-y^2\);}}}}{\color{#0066A7}{\uparrow }}} {=} 2\pi \dfrac{a}{b}\left( -\dfrac{1}{2}\right) \int_{b^{2}}^{0}\sqrt{u}\ {\it du} \\[-9pt] &=&\dfrac{\pi a}{b}\int_{0}^{b^{2}}u^{1/2}~{\it du}=\dfrac{\pi a}{b}\left[ \dfrac{ u^{3/2}}{\dfrac{3}{2}}\right] _{0}^{b^{2}}=\dfrac{2\pi a}{3b}( b^{3}) =\dfrac{2\pi ab^{2}}{3}\hbox{ cubic units} \end{eqnarray*}$$