Sphere volume using integrals

[Nickfytas]

My assignment is to Find the volume of the spherical cap in the figure below .using only double integral and polar coordinates. I tried many formulas but the one that must be correct is the one I write down below because for a=R I get half the volume of the sphere. I checked my calculations multiple times and I could find the correct answer which is given .please help

 

[Dr.Peterson]

Your integrand is z; but the volume you are looking for is not from the xy-plane to the sphere, but from the plane z=R - h to the sphere.

 

[Nickfytas]

if I understand correctly this is what you told me(in the photo)
if I have a circle in the xy axis and I sum all the circles from radious r=0 to r=a and angle θ=0 to Θ=2π shouldn’t I get the volume of the cup .
Also I’m confused what I’m I suppose to integrate .my text book says to integrate 1 to find the volume .and the limitations to calculate this only with polar coordinates confuses me even more .thank you for the help

 

[Dr.Peterson]

if I understand correctly this is what you told me(in the photo)
if I have a circle in the xy axis and I sum all the circles from radious r=0 to r=a and angle θ=0 to Θ=2π shouldn’t I get the volume of the cup .
Also I’m confused what I’m I suppose to integrate .my text book says to integrate 1 to find the volume .and the limitations to calculate this only with polar coordinates confuses me even more .thank you for the help

To find the volume, you integrate 1 in a triple integral, not in a double integral. That is, you are apparently thinking not of polar coordinates (for a plane), but of cylindrical coordinates (for space). In polar coordinates, you need to integrate the height (length of an element of volume, which is a vertical bar with height equal to the difference between the bottom and the top of the region).

My point was that the bottom of the region is not the xy plane, but the horizontal plane that is cutting the cap from the sphere.

In your new work there are several errors. You are still integrating z, as if the bottom of the region were the xy plane. In addition, you seem to be letting r vary from R-h to R; why would you do that? The integral itself is done wrong, because you didn't write out the substitution, and apparently failed to replace dr with its value in terms of u. (Your r, u, and h look similar, so I'm not sure I'm reading everything correctly.)

The only thing I said was wrong in your original work was the integrand! (I didn't look for errors beyond that.) Just subtract (R-h) from z to get the integrand.