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(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.

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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.

User Realph
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