# Thread: Integral of a hemisphere

1. ## Integral of a hemisphere

Hello everybody,

I am looking to solve a problem involving a triple (volume) integral. I intend to evaluate: $\iiint_{V}z(x^2+y^2+z^2)dxdydz$

where V is the upper hemisphere $0 \leq r \leq a$ and $z \geq 0$, with $r = \sqrt{x^2+y^2+z^2}$.

I wish to solve using both cartesian coordinates and then spherical polar coordinates, show that the result is equal. What I wish to ask of you is, how can I determine the limits of the integration for the cartesian coordinates. I don't know how I can differentiate between a hemisphere and cylinder in this respect, since in both cases the xy-plane is a circle and then the limits for z is just 0 to a.

But, to be honest, either way I don't think I can accurately determine the limits. I am having trouble picturing it. I know you have to take one variable at a time, and the limits will be a function of the 'remaining' variables, working down to no variables in the solution.

I have tried to rearrange $0 \leq r \leq a$ to get limits with one variable, say x, but then I end up with negative root on the left, which can't be right. So do I leave the lower limit as 0? Either way, I can't justify this. There may be just one simple idea I am missing.

Thank you

2. ## Re: Integral of a hemisphere

If this is the upper hemisphere, you would do better to integrate in the order "dzdydx" rather than "dxdydz". Using Cartesian coordiates, and itegrating in the order "dzdydx", the limits on the outer integral have to cover all possible x-value so run from -a to a. To determine the y limits, observe that the hemisphere projects down to the xy-axis as the region inside $x^2+ y^2= a^2$ or $y= \pm\sqrt{a^2- x^2}$ so y runs from $-\sqrt{a^2- y^2}$ to $\sqrt{a^2- x^2}$. Finally, since we have the upper hemisphere of $x^2+ y^2+ z^2= a^2$, which is equivalent to $z= \pm\sqrt{a^2- y^2- z^2}$, z will go from 0 to $\sqrt{a^2- x^2- y^2}$.

That is, the integral would be $\int_{x= -a}^a\int_{y= -\sqrt{a^2- x^2}}^{\sqrt{a^2- x^2}}\int_{z= 0}^{\sqrt{a^2- x^2- y^2}} z(x+ y+ z)dzdydx$.

If you want to use "spherical coorinates" then, at least, you need to understand what the coordinates mean! For any given point, (x, y, z), $\rho$ (not r) is the staright line distance from (0, 0, 0) to (x, y, z), $\theta$ is the angle, in the xy- plane, between the line from (0, 0) to (x, y) and the x-axis, so from 0 to $2\pi$ to cover the entire plane, and $\phi$ is the angle the line from (0, 0, 0) to (x, y, z) makes with the positive z-axis, so from 0 to $\pi$ to cover all the way from the positive z-axis to the negative z-axis.

So for the upper hemi-sphere, with center at (0, 0, 0) and radius a, $\rho$ goes from 0 to a, $\theta$ goes from 0 to $2\pi$, and $\phi$ goes from 0 to $\pi/2$.
$\int_{\phi= 0}^{\pi/2}\int_{\theta= 0}^{\pi/2}\int_{\rho= 0}^a \rho^2 cos(\phi)(cos(\theta)sin(\phi)+ sin(\theta)sin(\phi)+ cos(\phi))d\rho d\theta d\phi$.

3. ## Re: Integral of a hemisphere

Hello HallsofIvy,

Thank you for a great post!

I am aware the order the variables are integrated is changeable, I posted the problem as it was stated. I wasn't sure which was the best variable to start with. Looking at the limits you gave, I can see where I was going wrong in trying to comprehend it; though I did try to (unsuccessfully) visualise it as such. I can understand where these limits are coming from now, so thanks! So, for any upper hemisphere integral problem, these will remain the same, correct? (Although the 'a' could vary, I would simply substitute any given value in place for this). Regardless of the 'function' being integrated?

As for the spherical coordinates. I'll be honest and say that I had these limits in my notes! But, you have helped me to further understand the meaning, which will certainly help me to remember.

I will now solve using both coordinate systems and (hopefully) come up with the same answer in each case!

,

,

,

,

,

,

# what is the limit in hemisphere

Click on a term to search for related topics.