Solid by Revolution
Using the disk method is the easier method for this particular problem. Because a disk can be thought of as a cylinder and the equation for a cylinder is V=(pi)hr^2, if we were to think of the radius of the disk as the value of the function +7 and if the height of each disk was an infinitely small amount, we could piece together a function for each individual disk as V=(16-x^2)^2 dx.
Now if we were to take the integral of that function with the limits of integration as the points of intersection of the functions (-4 to 4), it would give us the volume of the solid.
The cylindrical shell method is a little more difficult than that however. To get a cylindrical shell, we would have to evaluate the integral as a function of y not a function of x. Therefore we should rearrange our parabola for y, which is y=sqrt(9-y). So now that we have that we need to think of the equation of the volume of a cylindrical shell, the volume being the circumference multiplied by the height and width. Because the distance between the large and small radii will be so insignificant, we can think of the width as dy. The equation for circumference is 2(pi)r, and because r is the distance between y=-7 and the function value, r can be thought of as (y+7). The height of each individual shell will be the value of x at each y value multiplied by two because it is a parabola and it's simetrical, so h=2sqrt(9-y).
So now if we put the two together, we get the equation V=2(pi)(y+7)2sqrt(9-y) dy
Now we integrate the function with the limits of integration being all the y-values between the line y=-7 and the maximum function value (-7 to 9) and that will give us the volume of the solid.
Closer to the end of class Mr. K gave us a stencil with similar problems on it.
(Sorry about my lack of the proper characters, the system I'm on at the moment has very limited resources.)