Volume of a solid of revolution: $y = x^3$, $y = x^{1/3}$, $x \geq 0$ rotated about $y$-axis
Sophia Terry 
I am trying to find the volume:
Rotate about $y$. $$y = x^3,\quad y = x^{1/3},\quad x \geq 0$$
Simple enough.
$x = y^3 \implies x = y^{\frac{1}{3}}$
$$\pi \cdot \int_0^1 y^{(1/3)^2} - y^{3^2}dy$$
$$\pi\cdot \left(\frac{3}{5} - \frac{1}{7}\right)$$
Of course that is wrong, why?
$\endgroup$ 61 Answer
$\begingroup$Edit: Since you're using the disk method, your solution is correct, given the problem as stated. But be careful with notation: note the difference between the squaring of the functions as shown below. You evaluated as it should be evaluated, so the result is correct.
$$\pi \cdot \int_0^1 \left(y^{1/3}\right)^2 - \left(y^{3}\right)^2\,dy = \pi \int_0^1 y^{2/3} - y^6\,dy$$
Which, integrating and evaluating, gives you
$$\pi\cdot \left(\frac{3}{5} - \frac{1}{7}\right) = \frac{16\pi}{35}$$
$\endgroup$ 3
