Triple Integral, Volume of a solid

[abbot]

Homework Statement

Well, first of all, I'm not english spoken, so sorry for the mistakes.
I was trying to calculate the integral below:

\int \int \int_{V} (xy+z) dxdydz

where V is a region in R^{3} bounded by

the sphere x^2+y^2+z^2<=9

the cone z^2<=x^2+y^2

and the plane z>=0


2. Relevant equations

The Attempt at a Solution

I tried to calculate it with spherical coordinates in that way:

x= r sin(β) cos(α)
y= r sin(β) sin(α)
z= r cos(β)

0<=α<=2π
π/4<=β<=π/2
0<=r<=3

but I'm not sure if that it's correct so can anybody help me?Thanks a lot

 

[Dick]

You have written z^2 LESS than or equal to x^2+y^2. If that's what you meant to write, I don't think the range of the beta angle is correct. Is it?

 

[abbot]

You have written z^2 LESS than or equal to x^2+y^2. If that's what you meant to write, I don't think the range of the beta angle is correct. Is it?

Well, I'm not sure, the solid bounded by those surfaces it's inside the sphere but outside the cone, so i thought beta angle goes from the straight lines that form cone, to the plane z=0, considering that beta angle starts at the z-axis

 

[Dick]

Well, I'm not sure, the solid bounded by those surfaces it's inside the sphere but outside the cone, so i thought beta angle goes from the straight lines that form cone, to the plane z=0, considering that beta angle starts at the z-axis

Ok, I'm just thinking of a different polar angle convention. I'd say that looks fine then.

 

[abbot]

I've also tried to calculate it integrating z in that way

\int \int( \int^{\sqrt{9-x^2-y^2}}_{\sqrt{x^2+y^2}} (xy+z) dz)dxdy

and then changing to polar coordinates with x and y
What do you think about it?

 

[Char. Limit]

I've also tried to calculate it integrating z in that way

\int \int( \int^{\sqrt{9-x^2-y^2}}_{\sqrt{x^2+y^2}} (xy+z) dz)dxdy

and then changing to polar coordinates with x and y

That's not what i would recommend. Change to spherical coordinates, then evaluate the triple integral. Merry Christmas!

 

[abbot]

That's not what i would recommend. Change to spherical coordinates, then evaluate the triple integral. Merry Christmas!

Thank you and Merry Christmas for you too!