Question

Theory and ExamplesVolume Find the volume of the solid generated by revolving the region enclosed by the ellipse 9x² + 4y² = 36 about the y−axis.

Accepted Answer

Rewrite the equation of the ellipse $$9x^{2}$$ + $$4y^{2} = 36 in $$terms of x to express x as a function of y. Start by isolating $$x^{2}$$: $$9x^{2} = 36$$ - $$4y^{2}$$, so $$x^{2}$$ = \frac{36 - $$4y^{2}$$}{9}. Then, x = \pm \sqrt{\frac{36 - $$4y^{2}$$}{9}} = \pm \frac{\sqrt{36 - $$4y^{2}$$}}{3}. Identify the limits of integration for y by finding the range of y-values on the ellipse. Since the ellipse is symmetric about the x-axis, find the maximum and minimum y by setting x=0 in the ellipse equation: $$9(0)^{2}$$ + $$4y^{2} = 36$$ \Rightarrow $$4y^{2} = 36$$ \Rightarrow $$y^{2} = 9$$ \Rightarrow y = \pm 3. So, y ranges from -3 to 3. Set up the volume integral using the method of cylindrical shells because the solid is generated by revolving around the y-axis. The formula for volume using shells is: $$V = \int_{a}^{b} 2\pi (	ext{radius})(	ext{height}) \, dy. $$Here, the radius is the distance from the y-axis, which is |x|, and the height is the thickness dy. Express the radius and height in terms of y. The radius is the x-value, which from step 1 is $$\frac{\sqrt{36 - 4y$$^{2}$$}}{3}. $$The height of each shell is the horizontal distance from -x to x, which is $$2x = 2 	imes \frac{\sqrt{36 - 4y$$^{2}$$}}{3}. $$However, since the shell method integrates with respect to y, the height corresponds to the horizontal length, so the height is $$2x$$ and the radius is $$|y| if $$revolving around y-axis, but since we revolve around y-axis, the radius is $$x$$ and height is dy. Actually, for revolving around y-axis, the shell radius is $$|x|$$ and height is the vertical segment in y, so we should integrate with respect to x. Alternatively, use the disk/washer method integrating with respect to y. Use the disk/washer method integrating with respect to y. The radius of each disk is the x-value from the ellipse, $$R(y) = \frac{\sqrt{36 - 4y$$^{2}$$}}{3}. $$The volume is then $$V = \pi \int_{-3}^{3} [R(y)]^{2} dy = \pi \int_{-3}^{3} \left( \frac{\sqrt{36 - 4y$$^{2}$$}}{3} \$$right)^{2}$$ dy. $$Simplify the integrand and set up the integral to find the volume.

Theory and Examples

Volume Find the volume of the solid generated by revolving the region enclosed by the ellipse 9x² + 4y² = 36 about the y−axis.

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Key Concepts

Equation of an Ellipse

Method of Cylindrical Shells

Volume of Solids of Revolution