Using triple integral, I need to find the volume of the solid region in the first octant enclosed by the circular cylinder r=2, bounded above by z = 13 - r^2 a circular paraboloid, and bounded below by the (x,y) plane. I need to sketch this as well. If you could please help me, that would be much appreciated.

The integral  int_0^2(13sqrt(4-x^2)-x^2 sqrt(4-x^2)-1/3(4-x^2)^(3/2))dx.
Perform a substitution x=2sin u, then dx = 2cos u du, 4-x^2=4cos^2 u and u changes from 0 to pi/2:
int_0^(pi/2) (13*2cos u-4sin^2 u*2cos u-1/3*8cos^3 u)*2cos u du =


= 2 int_0^(pi/2) (18cos^2 u+16/3 cos^4 u) du.
 
Now use the formula cos^2 y = 1/2(1+cos(2y)) several times:
18cos^2 u+16/3 cos^4 u = 9(1+cos(2u))+4/3(1+cos(2u))^2=
=31/3+35/3cos(2u)+4/3 cos^2(2u) =31/3+35/3cos(2u)+2/3+2/3cos(4u) = 11 + 35/3cos(2u)+2/3 cos(4u).
It is clear that the integrals of cos(2u) and of cos(4u) are zeros and the final answer is  2int_0^(pi/2) 11 du = 11pi.
 

Hello! As I understand, r = sqrt(x^2 + y^2).
The volume is the integral by dx, dy and dz of the function that is constantly equal to 1. To set up the integral we need to understand the limits for each variable (maybe for given values of other variables).
Let's start from the x variable. It may be between 0 and 2. For any given x, the variable y can be between 0 and sqrt(4-x^2). And for given x and y the variable z can be between 0 and 13-x^2-y^2.
This way the desired triple integral is
V = int_(x=0)^(x=2) dx int_(y=0)^(y=sqrt(4-x^2)) dy int_(z=0)^(z=13-x^2-y^2) dz (1).
 
We can now solve the integrals starting from the innermost:
int_(z=0)^(z=13-x^2-y^2) dz (1) =13-x^2-y^2,
int_(y=0)^(y=sqrt(4-x^2)) (13-x^2-y^2) dy = 13sqrt(4-x^2) - x^2sqrt(4-x^2) - 1/3 (4-x^2)^(3/2).
The third integral,  int_0^2 (13sqrt(4-x^2) - x^2sqrt(4-x^2) - 1/3(4-x^2)^(3/2)) dx,
is not so simple but I believe the main task is to deal with triple integral. Integrating all terms starts from the trigonometric substitution x=2sin(u).
The result of indefinite integration is  22arcsin(x/2)-1/6xsqrt(4-x^2)(x^2-37)+C,  the definite integral is 11 pi.
It is the volume we had to find.
Please look at the picture. Please tell me if you need the explicit integration of the dx integral and I'll add it.