# Find the exact area of the surface obtained by rotating the given curve about the x-axis. x =...

## Question:

Find the exact area of the surface obtained by rotating the given curve about the {eq}x {/eq}-axis.

{eq}x = a \cos^3(\theta), \; y = a \sin^3(\theta), \; 0 \leq \theta \leq \frac{\pi}{2} {/eq}

## Surface Area:

To find the surface of parametric curve the following formula is used {eq}\int \sqrt{\left ( \frac{\mathrm{d} x}{\mathrm{d} \theta} \right )^{2}+\left ( \frac{\mathrm{d} y}{\mathrm{d} \theta} \right )^{2}}d\theta {/eq} where {eq}\theta {/eq} lies from 0 to {eq}\frac{\pi}{2} {/eq} as given in the problem.

## Answer and Explanation:

{eq}\int_{0}^{\frac{\pi}{2} } \sqrt{\left ( \frac{\mathrm{d} x}{\mathrm{d} \theta} \right )^{2}+\left ( \frac{\mathrm{d} y}{\mathrm{d} \theta} \right )^{2}}d\theta\\ x=a\cos^{3}\theta\\ \frac{\mathrm{d} x}{\mathrm{d} \theta}=-3a\cos^{2}\theta \sin\theta\\ y=a\sin^{3}\theta\\ \frac{\mathrm{d} x}{\mathrm{d} \theta}=3a\sin^{2}\theta \cos\theta\\ =\int_{0}^{\frac{\pi}{2} } \sqrt{\left ( -3a\cos^{2}\theta \sin\theta\right )^{2}+\left ( 3a\sin^{2}\theta \cos\theta \right )^{2}}d\theta\\ =\int_{0}^{\frac{\pi}{2} } 3a\cos\theta\sin\theta d\theta\\ =\frac{3a}{2} {/eq}

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