# 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...

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.

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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.

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.`