## University Calculus: Early Transcendentals (3rd Edition)

a) $$\int^1_0r\sqrt{1-r^2}dr=\frac{1}{3}$$ b) $$\int^1_{-1}r\sqrt{1-r^2}dr=0$$
Part a) $$A=\int^1_0r\sqrt{1-r^2}dr$$ Set $u=1-r^2$, we then have $$du=-2rdr$$ $$rdr=-\frac{1}{2}du$$ For $r=0$, we have $u=1$ and for $r=1$, we have $u=0$. Therefore, $$A=-\frac{1}{2}\int^0_1\sqrt udu=-\frac{1}{2}\int^0_1u^{1/2}du$$ $$A=-\frac{1}{2}\times\frac{2u^{3/2}}{3}\Big]^0_1=-\frac{u^{3/2}}{3}\Big]^0_1$$ $$A=-\frac{1}{3}(0-1)=\frac{1}{3}$$ Part b) $$A=\int^1_{-1}r\sqrt{1-r^2}dr$$ We carry out a similar substitution as in part a). For $r=-1$, we have $u=0$ and for $r=1$, we have $u=0$. Therefore, $$A=-\frac{1}{2}\int^0_0\sqrt udu=0$$