# Volume inside a hemisphere and a cylinder

1. Nov 9, 2015

### qq545282501

1. The problem statement, all variables and given/known data
use spherical coordinates to find the volume of the solid inside the hemisphere z= √(25-x^2-y^2) and bounded laterally by the cylinder x^2+y^2=4

2. Relevant equations
x=rcosθ =ρsinφcosθ , y=rsinθ =ρsinφsinθ
z=ρcosφ
r= ρsinφ

3. The attempt at a solution
I divided the solid into 2 parts, upper part is the small dome and below it is a cylinder with radius of 2 and height of sqrt(21)[ by setting z=sqrt(25-4) ]

so the volume of the cylinder = 2(r^2)h=57.59
volume of the dome= 328.98
adding the 2 volumes together give 396.87. which is greater than the volume of a hemisphere of radius 5.
so this is very wrong, but i dont know where is my mistake

2. Nov 9, 2015

### SteamKing

Staff Emeritus
That's not what this equation of the cylinder means. The cylinder has an infinite extent in the z-direction. The axis of the cylinder is coincident with the z-axis and the cylinder has a radius of 2. The hemisphere is centered at the origin, has a radius of 5, and encloses its volume on the positive z side of the x-y plane.

You should make a sketch before diving into writing the integral.

You want to calculate the volume of the cylinder between the x-y plane and the surface of the hemisphere. The cylinder has a height of $\sqrt{21}$ only on its central axis; all other heights are shorter.

The volume of a cylinder is $πr^2h$, but the h varies in this case, due to the location of the surface of the hemisphere which is intersected by the cylinder.
The entire volume of the hemisphere is only $\frac{2}{3}πr^3=261.8$, so I don't know how you calculated 329.98.

3. Nov 9, 2015

### qq545282501

I dont understand, was my sketch incorrect? 1+2= the volume? am i not correct on this?

4. Nov 9, 2015

### SteamKing

Staff Emeritus
Your sketch looks correct, but something went drastically wrong in the arithmetic of the volume computation.

The volume you calculated for the cylinder, item 2, or 57.59, is correct. Your calculation of the volume of the spherical cap above the cylinder, item 1, is way off.

You still have to find a way of expressing the volume of items 1 and 2 in terms of an integral using spherical coordinates. It might be less complicated if you took advantage of symmetry here, since the cylinder and the hemisphere are concentric with one another.

Also, since you are supposed to use spherical coordinates, perhaps splitting the volume into items 1 and 2 as shown might not be the least complicated method here. I suggest drawing a radius from the center of the hemisphere to the intersection it makes with the cylinder. You would have a cone with a partly spherical cap, roughly shaped like an ice cream cone. The rest of the volume could be obtained by revolving the triangle below the ice cream cone extending to the side of the cylinder around the z-axis and then adding it to the volume of the ice cream cone.

5. Nov 9, 2015

### qq545282501

are my limits on the dome volume correct? if not, what should be the limits for the rho? I understand your method, but doesn't it requires 2 integrals, making it harder? i mean the ice cream cone is 1 integral and revolving the triangle below the ice cream around z-axis also requires another integral. well, i am gona give it a try using your method.
V1= volume of the ice cream cone= 12.2339
V2 is where i am having trouble with:

V2=

i know V2 is wrong, wolfram wont even compute it.

Last edited: Nov 9, 2015
6. Nov 9, 2015

### qq545282501

I found another way to solve the problem. thx