I'm trying to determine the volume of the solid that is obtained by rotating around the $x$-axis the region limited by $y=2x$ and $y=x^2$. I'm trying to follow James Stewart's "Calculus. Single variable" book, but I'm not fully understanding. Is the area given by
$$
A=\int_0^2[\pi(2x-x^2)^2]dx
$$
Or should I calculate the area between these regions:
$$
A=\int(2x-x^2)dx=x^2-\frac{x^3}{3}
$$
And then the volume as:
$$
V=\int_0^2{[\pi(x^2-\frac{x^3}{3})^2]dx}
$$
I'm a bit confused here, any help will be much appreciated.
Best Answer
For determining the volume of the solid that is obtained by rotating around the $x$-axis limited by $f(x)$ and $g(x)$ where $f(x)\geq g(x)$ in $[a,b]$ you have $$V=\pi\int_a^b f^2(x)-g^2(x)\ dx$$ here $$V=\pi\int_0^2 4x^2-x^4\ dx$$