Answer
a)$\frac{1}{3}$
b)0
Work Step by Step
a)$Let u=1-r^2$=>du=-2rdr
=>$\frac{-1}{2}du=rdr;$
r=0=>u=1,r=1=>u=0
$\int^{1}_{0}r\sqrt{1-r^2}dr$
=$\int^0_1 \frac{-1}{2}\sqrt{u}$
=$[\frac{-1}{3}u^{3/2}]^0_1$
=$0-(\frac{-1}{3})(1)^{3/2}$
=$\frac{1}{3}$
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b) Use the same substituition for u as in part a
r=-1=>u=0
r=1=>u=0
$\int ^1_{-1}r\sqrt{1-r^2}dr$
=$\int^0_0\frac{-1}{2}\sqrt{u}du$
=0
