## Calculus: Early Transcendentals 8th Edition

The area of the surface obtained by rotating the curve about the y-axis is: \begin{aligned} S&=\int 2 \pi x d s \\ &=\int^{\frac{a}{2}}_{0} 2 \pi (\sqrt{a^{2}-y^{2}} ) \sqrt {\left(\frac{a^{2}}{{a^{2}-y^{2}}}\right)} dy \\ &= \pi a^{2} \end{aligned}
$$x= \sqrt{a^{2}-y^{2}}, \quad \quad 0 \leq y \leq \frac{a}{2}$$ $\Rightarrow$ $$\frac{dx}{dy}= \frac{-y}{\sqrt{a^{2}-y^{2}}}$$ $\Rightarrow$ \begin{aligned} ds &= \sqrt {1+\left(\frac{dx}{dy}\right)^{2}} dy \\ &=\sqrt {1+\left(\frac{-y}{\sqrt{a^{2}-y^{2}}}\right)^{2}} dy\\ &=\sqrt {1+\left(\frac{y^{2}}{{a^{2}-y^{2}}}\right)} dy \\ &=\sqrt {\left(\frac{a^{2}}{{a^{2}-y^{2}}}\right)} dy \end{aligned} So, the area of the surface obtained by rotating the curve about the y-axis is: \begin{aligned} S&=\int 2 \pi x d s \\ &=\int^{\frac{a}{2}}_{0} 2 \pi (\sqrt{a^{2}-y^{2}} ) \sqrt {\left(\frac{a^{2}}{{a^{2}-y^{2}}}\right)} dy \\ &=\int^{\frac{a}{2}}_{0} 2 \pi (\sqrt{a^{2}-y^{2}} )\frac{a}{ \sqrt {a^{2}-y^{2}}} dy \\ &=2 \pi \int^{\frac{a}{2}}_{0} a dy\\ &=2 \pi a\left[y\right]_{0}^{\frac{a}{2}}\\ &=2 \pi a\left[ \frac{a}{2}\right]\\ &= \pi a^{2} \end{aligned}