Its calculus but it is also 30 !

write the integral in one variable to find the volume of the solid obtained by rotating the first-quadrant region bounded by y = 0.5x^2 and y = x about the line x = 7

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).

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.

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.

To solve this integral, what you can do is split it up as follows: (1) ∫40(x2−4)dx=∫40(x2)dx−∫40(4)dx. to integrate  4  as an  indefinite integral, a good rule of thumb is  by  an equation below  for  polynomials; . (2) ∫xa=xa+1a+1+c , where a is any number. c denotes some constant that is crucial when doing