# Trouble understanding meaning of triple integral in spherical coordinates

• zjmarlow
On the other hand, if f(x,y,z) is a scalar function, then the integral doesn't have any physical interpretation at all!In summary, the integral \displaystyle \iiint\limits_B e^{x^2 + y^2 + z ^2}dV evaluates the volume of the function \displaystyle e^{x^2 + y^2 + z ^2} within the unit ball. It does not have a specific physical interpretation and is not four-dimensional as it is integrating over three independent variables. The function inside the integral can have different meanings depending on the application, such as representing density or being a scalar function.

## Homework Statement

Evaluate $$\iiint\limits_B e^{x^2 + y^2 + z ^2}dV$$ where $$B$$ is the unit ball.

See above.

## The Attempt at a Solution

Does this evaluate the volume of $f(x, y, z)$ within the unit ball (i.e. anything falling outside the unit ball is discarded)? Also, since $f$ is a function of three independent variables, I'm confused - wouldn't that be four dimensional?

Last edited:
zjmarlow said:

## Homework Statement

Evaluate $$\iiint\limits_B e^{x^2 + y^2 + z ^2}dV$$ where $$B$$ is the unit ball.

See above.

## The Attempt at a Solution

Does this evaluate the volume of $$f(x, y, z)$$ within the unit ball (i.e. anything falling outside the unit ball is discarded)? Also, since $$f$$ is a function of three independent variables, I'm confused - wouldn't that be four dimensional?
Yes, you are integrating the function, $\displaystyle e^{x^2 + y^2 + z ^2}$ over the volume of a unit ball.

Remember that $\displaystyle x^2 + y^2 + z ^2=\rho^2\ .$

Thank you for your reply. The solution is worked out: $\frac{4\pi}{3}(e - 1)$. My problem now is visualizing $e^{x^2 + y^2 + z^2}$. Could you explain what this would look like?

Or is $\frac{4\pi}{3}(e - 1)$ like a scalar field that says how much each position in the unit ball is "worth"?

zjmarlow said:
Or is $\frac{4\pi}{3}(e - 1)$ like a scalar field that says how much each position in the unit ball is "worth"?

How did you arrive at that result?

I looks as if you have you have made some gross simplification.

Sorry, that should be $f(x, y, z)$ (in general) or $e^{x^2 + y^2 + z^2}$ (in this particular case).

Show what integral you have in spherical coordinates.

Ok, I copied the example problem wrong. It should be
$$\iiint\limits_B e^{(x^2 + y^2 + z^2)^{(3/2)}}dV$$
which becomes
$$\int_0^{2\pi} \int_0^\pi \int_0^1 e^{(\rho^2)^{(3/2)}} \cdot \rho^2 sin \phi d\rho d\phi d\theta$$
$$= \int_0^{2\pi} \int_0^\pi \frac{e - 1}{3} sin \phi d\phi d\theta$$
$$= \int_0^{2\pi} \frac{2}{3}(e - 1) d\theta$$
$$= \frac{4\pi}{3}(e - 1)$$

I guess I see how the integral without the function calculates volume (the book has a nice picture of a curved box) but what's the meaning of the function you integrate? What does something like $e^{(x^2 + y^2 + z^2)^{(3/2)}}$ or $e^{\rho^3}$ even look like?

zjmarlow said:
Ok, I copied the example problem wrong. It should be
$$\iiint\limits_B e^{(x^2 + y^2 + z^2)^{(3/2)}}dV$$
which becomes
...

I guess I see how the integral without the function calculates volume (the book has a nice picture of a curved box) but what's the meaning of the function you integrate? What does something like $e^{(x^2 + y^2 + z^2)^{(3/2)}}$ or $e^{\rho^3}$ even look like?
Well, that makes more sense.

The integral $\displaystyle \int e^{\rho^3}\rho^2 \,d\rho$ is pretty straight forward.

The integral $\displaystyle \int e^{\rho^2}\rho^2 \,d\rho$ involves the error function.

If, as before, f(x,y,z) is density, measured in kilograms per cubic meter, then the integral will have units of $(kg/m^3)(m^3)= kg$