Washer method:
The volume of an arbitrary washer is:
\(\displaystyle dV=\pi(R^2-r^2)\,dx\)
The outer radius R is constant; it is the distance between the lower boundary of the rotated region, \(\displaystyle y=2\) and the axis of rotation \(\displaystyle y=-4\):
\(\displaystyle R=2-(-4)=2+4=6\)
The inner radius is the distance between the upper boundary of the rotated region, \(\displaystyle y=-x^2\) and the axis of rotation \(\displaystyle y=-4\):
\(\displaystyle r=2-(-x^2)=x^2+2\)
and so we have:
\(\displaystyle dV=\pi(6^2-(x^2+2)^2)\,dx=\pi(-x^4-4x^2+32)\,dx\)
Adding up the washers, we find:
\(\displaystyle \displaystyle V=\pi\int_{-2}^0(-x^4-4x^2+32)\,dx\)
If the integrand is an even function, i.e., \(\displaystyle f(-x)=f(x)\) then we have:
\(\displaystyle \displaystyle \int_{-b}^0f(x)\,dx=\int_0^b f(x)\,dx\)
which we can demonstrate with a substitution:
Let \(\displaystyle u=-x\,\therefore\,du=-dx\) and the left side becomes:
\(\displaystyle \displaystyle -\int_{b}^0f(-u)\,du=\int_0^b f(x)\,dx\)
Now, the anti-derivative form of the fundamental theorem of calculus gives us:
\(\displaystyle \displaystyle -\int_{b}^af(x)\,dx=\int_a^b f(x)\,dx\)
\(\displaystyle -(F(a)-F(b))=F(b)-F(a)\)
\(\displaystyle F(b)-F(a)=F(b)-F(a)\)
and so we have:
\(\displaystyle \displaystyle \int_0^b f(-u)\,du=\int_0^b f(x)\,dx\)
Now, since we have assumed \(\displaystyle f(x)\) is even, then we have:
\(\displaystyle \displaystyle \int_0^b f(u)\,du=\int_0^b f(x)\,dx\)
And finally switching the dummy variable of integration, we obtain:
\(\displaystyle \displaystyle \int_0^b f(x)\,dx=\int_0^b f(x)\,dx\)
Now, this means we may write the volume as:
\(\displaystyle \displaystyle V=\pi\int_0^2(-x^4-4x^2+32)\,dx\)
We could have easily evaluated the volume in its original form, I just wanted to demonstrate a use of symmetry you may find useful in the future.
You should verify that:
\(\displaystyle V=\dfrac{704\pi}{15}\)
Now let's look at the shell method. The volume of an arbitrary shell is:
\(\displaystyle dV=2\pi rh\,dy\)
The radius of the shell is the distance betwee the axis of rotation and the variable of integration y:
\(\displaystyle r=2-y\)
The height of the arbitrary shell is equal to the distance between the y-axis and x.
\(\displaystyle h=|0-x|=|x|=\sqrt{x^2}=\sqrt{-y}\)
and so we have:
\(\displaystyle dV=2\pi(2-y)\sqrt{-y}\,dy\)
Summing up the shells we have:
\(\displaystyle \displaystyle V=2\pi\int_{-4}^0(2-y)\sqrt{-y}\,dy\)
Let \(\displaystyle u=-y\,\therefore\,du=-dy\) and we have:
\(\displaystyle \displaystyle V=-2\pi\int_4^0(2+u)\sqrt{u}\,dy=2\pi\int_0^4\left(2u^{ \frac{1}{2}}+u^{\frac{3}{2}} \right)\,du\)
You should verify that:
\(\displaystyle V=\dfrac{704\pi}{15}\)
Why use both methods? Two reasons: it gives you a means of checking your result and it is simply good practice to use both methods, at least when it is practical to do so.
