Learn How to Evaluate Triple Integrals: Volume vs Surface Integral Method

[jlmac2001]

Evaluate the triple integral (grad dot F dV) over the region x^2+y^2+z^2 less than or equal to 25 if F+(x^2+y^2+z^2)(xi+yj+zk), by doing either a volume or a surface integral, whichever is easiest!


will each of the integral go from -5 to 5? what is the easiest way?

 

[HallsofIvy]

You have a problem:

If you are attempting a problem like this, you should already have learned how to set up a general triple integral!

No, the three integrals do not all go from -5 to 5. That would be a rectangular solid. Your figure here is a sphere. If you wanted to integrate over the volume of the sphere, in cartesian coordinates, you would have, say, x form -5 to 5, y from -√(25-x²) to +√(25-x²), z from -√(25-x²-y²) to +√(25-x²-y²).
You could also use spherical coordinates in which you would have ρ from 0 to 5, θ from 0 to 2π, φ from 0 to π or cylindrical coordinates: r from 0 to 5, θ from 0 to 2π, z from -√(25- r²) to √(25- r²).

In this case, with F= (x^2+y^2+z^2)(xi+yj+zk), &del;.F (That is, by the way "del dot F", not "grad dot F": "grad" is specifically del of a scalar function), also called "div F" or the "divergence of F" is complicated. I would be inclined to use the divergence theorem: $$\int\int_T\int(\del . v)dV= \int_S (v.n)dS$$ where T is a three dimensional solid, S is the surface of the solid, and n is the unit normal vector to the surface at each point.

Since the surface of the figure is given by x²+ y²+z²= 25, The (not-unit) normal vector is given by the gradient of the left hand side: 2xi+ 2yj+ 2zk. If we choose to do the integration in the xy-plane, we divide by the k coefficient:
n dσ= ((x/z)i+ (y/z)j+ k)dydx. and integrate on the circle x²+y²= 25.

I would be inclined to use polar coordinates to do that since, then, z= √(25- r²) and F= 25(rcosθi+rsinθj+ &radic(25- r²)k)
(In order to get the entire surface of the sphere you would need to do z>0 and z< 0 separately. Alternatively, because of the symmetry, just multiply by 2.)

 

[M98Ranger]

The integral will indeed go from -5 to 5 for all three variables, since the region x^2+y^2+z^2 less than or equal to 25 is a sphere with radius 5 centered at the origin.

To evaluate this triple integral, we can use either the volume or surface integral method. The volume integral method involves splitting the region into small infinitesimal cubes and calculating the contribution of each cube to the overall integral, while the surface integral method involves evaluating the integral over the surface of the region.

In this case, since the function F is in terms of x^2+y^2+z^2, it would be easier to use the volume integral method. This is because the region is a sphere, so splitting it into small cubes would result in each cube having the same contribution to the integral. We can then use the formula for the volume of a sphere to calculate the contribution of each cube, and then integrate over all the cubes to get the final answer.

On the other hand, if the function F was not in terms of x^2+y^2+z^2, it might be easier to use the surface integral method. This is because the surface integral method allows us to directly integrate over the surface of the region, which might be more straightforward than calculating the contribution of each cube.

In general, the choice between volume and surface integral method depends on the specific function and region involved. It is important to understand both methods and choose the one that is easier for a given problem.