Question

(1 point) The volume of the solid obtained by rotating the region enclosed by x=2y,y3=x(with y≥0) about the y-axis can be computed using the method of disks or washers via an integral V=∫ba ? with limits of integration a= and b= . The volume is V= cubic units. Note: You can earn full credit if the last question is correct and all other questions are either blank or correct.

Answer 1

$x=2y$ and $x=y^3$ intersect when

$2y=y^3\implies y^3-2y=y(y^2-2)=y(y-\sqrt2)(y+\sqrt2)=0$

$\implies y=0\text{ or }y=\sqrt2\text{ or }y=-\sqrt2$

We omit the negative root, since we only care for $y\ge0$.

In the interval (0, √2), we have $y^3<2y$. Then the volume is given by the integral

$\displaystyle\pi\int_0^{\sqrt2}(2y)^2-(y^3)^2\,\mathrm dy=\pi\int_0^{\sqrt2}4y^2-y^6\,\mathrm dy$

and so the volume is (32√2)/21 π.