$y = x^3, y=8, x=0$ rotated about $x$-axis
so volume for cylindrical shells is = to the integral of (circumfrence)(height)(thickness), correct?
so I have $2\pi \int_0^8 y(8-y^{1/3} ) dy = 2\pi \int_0^8 8y - y^{4/3} dy = (4y^2 - (3/7) y^{7/3})|_0^8 = 4*64-(3/7)*128$
I get $201.14xxxx$ times $2\pi$ which is $(1408*2\pi)/7 $
my book says the answer is $ 768\pi/7 $
I don't get what I could have done wrong here? does anyone see the flaw in my math??
I like how at least the denominator of my answer and the books answer is the same, that must mean im doing some calculation right, but I can't pinpoint where I'm wrong.
You had set up the wrong integral.
The circumference is given by $2\pi y$, the width by $y^{\frac{1}{3}}$ and the thickness is $dy$ so your integral should be:
$$\int_0^82\pi y^{\frac{4}{3}} dy$$
$$=2\pi\left(\frac{3}{7}y^\frac{7}{3}\right)_0^8$$
$$=2\pi\cdot\frac{3}{7}\cdot8^\frac{7}{3}$$
$$=2\pi\cdot\frac{3}{7}\cdot128$$
$$=\frac{768\pi}{7}$$