To solve this, we are going to be revolving this area around the line x=7, which means that we are rotating with respect to y, meaning we must change the equations to x= f(y).
[tex]y = .5x^2
\\ 2y=x^2
\\ \sqrt{2y} =x
\\ x = y [/tex]
Now that we have those two equations, we can graph this (I'm going to assume this is a calculator FRQ/MC problem) and find our top and bottom bounds. Looking at the graph, we can see that the bounds are 0 and 2. Now comes the part in which we set up the equation of the integral.
[tex]\pi \int\limits^2_0 {((y-7)^2- (\sqrt{2y}-7)^2) } \, dy[/tex]
Because this area is being revolved around the line x=7, we can set up an equation with this in mind. We use y-7 in order to take into account the axis of rotation, and the equation of the line x=y, and the same with the other function. We put y-7 first because it is the outer function in relation to the function y=.5x^2. Now we integrate (and once again, assuming that this is a calculator FRQ/MC) and get 25.1327.
Please, please, please let me know if this is incorrect, because I'm a bit hazy on this as well, I'll make sure to correct it as soon as possible. Thanks.
