Find the volume generated by revolving the area bounded by $y^2=x^3$ , $x=4$ about the line $x=1$.
I can't understand how the area will revolve about a line lying on the area. Many thanks in advance.
Find the volume generated by revolving the area bounded by $y^2=x^3$ , $x=4$ about the line $x=1$.
I can't understand how the area will revolve about a line lying on the area. Many thanks in advance.
The volume of a solid of revolution is the integral of the square of the distance from the axis of revolution times $\frac{\tau}2=\pi$. The part of the curve to the left of $x=1$, where $-1
volume $=\frac{\tau}2 (\int_{1}^{8}( \ ^3\sqrt{y^2}-1)^2 \ dy+\int_{-8}^{1}( \ ^3\sqrt{y^2}-1)^2 \ dy)$
$\int (^3\sqrt{y^2}-1)^2dy=\int y^{\frac{4}3}-2y^{\frac{2}3}+1 \ dy=\frac{3}7 y^{\frac{7}3}-\frac{6}5 y^{\frac{5}3}+y$