# volume problem - cylindrical shell method

• Jul 20th 2009, 04:31 PM
superdude
volume problem - cylindrical shell method
"The region enclosed by the curves $y=x^{3}$ and $y=\sqrt{x}$ is roated about the line $x=1$. Fint the volume of the resulting solid" -James Stewart

answer: $\frac{13\pi}{30}$
what I did:
difference of shells formula is $2\pi \int x(f(x)-g(x))$
so in this problem $f(x)=\sqrt{x}$ and $g(x)=x^3$. The intersections are $x=0,1$. I get confused because it's not being rotated around the y-axis but $x=1$
I tried $V= 2 \pi \int_0^1 x((1-\sqrt{x})-(1-x^{3})$ but I end up with a negative number.
• Jul 20th 2009, 05:57 PM
Calculus26
The x term represents the distance between an arbitrary x and the axis of rotation.

If the solid is rotated about the yaxis the x term is just x.

In your example the distance between x and the axis of rotation x = 1 is

1-x. so replace x with 1-x--- you obtain (1-x)*(sqrt(x)-x^3) for the integrand.

A more general form of the shell method is 2pint{D(x)*(f-g)dx}

Where D(x) is as mentioned the distance between an arbitrary x and the axis of rotation