Double integral and polar cordinates other problem.

[christian0710]

If we have to find the volume, written in polar cordinates, inside this sphere X²+y²+z²=16 and outside this cylinder x²+y²=4

How should I approach this?
Could I take the sphere function and reqrite in polar cordinates z=√(16-X²-y²) which is the same as z=√(16-r²)

But then I have to make r depend on a function of the cylinder right?

so x²+y²=4 ---> r²=4 so r=2 this must be the boundaries of the cylinder..

Now I get a bit confused. Do we subtract the sphere function from the cylinder function? Or do we make the cylinder function a function that r depends on?

 

[SammyS]

If we have to find the volume, written in polar cordinates, inside this sphere X²+y²+z²=16 and outside this cylinder x²+y²=4

How should I approach this?
Could I take the sphere function and reqrite in polar cordinates z=√(16-X²-y²) which is the same as z=√(16-r²)

But then I have to make r depend on a function of the cylinder right?

so x²+y²=4 ---> r²=4 so r=2 this must be the boundaries of the cylinder..

Now I get a bit confused. Do we subtract the sphere function from the cylinder function? Or do we make the cylinder function a function that r depends on?

What does r have to be to be outside the cylinder?

 

[christian0710]

Hmm then this would make sense 2<r<4 because the cylinder has a radius of 2, so i guess the theta would go from 0 to 2pi ?

 

[SammyS]

Hmm then this would make sense 2<r<4 because the cylinder has a radius of 2, so i guess the theta would go from 0 to 2pi ?

Yes.

What is the function you integrate, i.e., what is your integrand?

 

[christian0710]

The function we want to integrate must be the sphere, right?
X^2+y^2+z^2=16 so if we isolate z we must get z=√(16-X^2-y^2) and written in polar form
z=√(16-r^2) So this is the function we integrate?

 

[SammyS]

The function we want to integrate must be the sphere, right?
X^2+y^2+z^2=16 so if we isolate z we must get z=√(16-X^2-y^2) and written in polar form
z=√(16-r^2) So this is the function we integrate?

That's the equation for the top half of the sphere, the half for which z is positive.

 

[christian0710]

I see, so if 2<r<4 and 0< theta< 2pi
Hmm so we could add two double integrals? one of the lower, and one of the upper sphere?

 

[SammyS]

I see, so if 2<r<4 and 0< theta< 2pi
Hmm so we could add two double integrals? one of the lower, and one of the upper sphere?

... or multiply the first integral by 2, because of the symmetry involved.

Do you know what dxdy , the element of area, is in polar coordinates?

 

[christian0710]

yes that's a smart solution :) The polar form of dxdy must be r*dr*dθ Right?

I assume it means the rectangles delta r times delta θ multiplied by the length r (a bit hard for me to visualize)

 

[christian0710]

So 2*∫(∫((√(16-r^2) )*r,r,2,4),θ,0,2∏) This must be the solution? :)

 

[SammyS]

2*∫∫(√(16-r^2) )*r*dr*dθ; where r goes from 2 to 4, θ goes from 0 to 2π .

In my opinion, you should always include the differentials when writing an integral.

 

[christian0710]

Ahh yes a good idea. Thank you very much for your help :)

 

[christian0710]

Hmm but is there an easier way? Because (√16-r²)r is not that easy to do by hand, and id love to be able to do this by hand :)

 

[SammyS]

Hmm but is there an easier way? Because (√16-r²)r is not that easy to do by hand, and id love to be able to do this by hand :)

[STRIKE]Multiply through by r .[/STRIKE]

Added in Edit:

Putting the parentheses in the correct place gives

( √(16-r²) )r​

Do integration by substitution.

What is the derivative of 16-r² ?