Mathematics

Problem

Let

R

be the region enclosed by

x=4-y

and

x=y^{2}

for

0\le y\le\dfrac{-1+\sqrt{17}}{2}

. When

R

is revolved about the vertical line

x=-1

, which definite integral represents the volume of the resulting solid of revolution? a)

\displaystyle \pi\int_{0}^{\frac{-1+\sqrt{17}}{2}}\Bigl[(5-y)^{2}-(y^{2})^{2}\Bigr]dy

\displaystyle \pi\int_{0}^{\frac{-1+\sqrt{17}}{2}}(5-y)^{2}\,dy

\displaystyle \pi\int_{0}^{\frac{-1+\sqrt{17}}{2}}\Bigl[(5-y)^{2}-(y^{2}+1)^{2}\Bigr]dy

\displaystyle \pi\int_{0}^{\frac{-1+\sqrt{17}}{2}}\Bigl[(y^{2}+1)^{2}-(5-y)^{2}\Bigr]dy