Answer
$V=\pi(\ln4-\frac{1}{2})$
Work Step by Step
$A(y)=\pi(1+\frac{1}{y})^2-\pi(1+1)^2$
$A(y)=\pi(2y^{-1}+y^{-2}-3)$
$V=\int_{\frac{1}{2}}^{1}A(y)dy$
$V=\int_{\frac{1}{2}}^{1}\pi(2y^{-1}+y^{-2}-3)dy$
$V=\pi\int_{\frac{1}{2}}^{1}(2y^{-1}+y^{-2}-3)dy$
$V=\pi[2\ln{y}-y^{-1}-3y]_{\frac{1}{2}}^{1}$
$V=\pi((2\ln1-1^{-1}-3(1))-(2\ln{\frac{1}{2}}-\frac{1}{2}^{-1}-3(\frac{1}{2})))$
$V=\pi(\ln4-\frac{1}{2})$
