# Homework Help: Help with integration

1. Feb 10, 2004

### 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?

2. Feb 11, 2004

### 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 -&radic;(25-x2) to +&radic;(25-x2), z from -&radic;(25-x2-y2) to +&radic;(25-x2-y2).
You could also use spherical coordinates in which you would have &rho; from 0 to 5, &theta; from 0 to 2&pi;, &phi; from 0 to &pi; or cylindrical coordinates: r from 0 to 5, &theta; from 0 to 2&pi;, z from -&radic;(25- r2) to &radic;(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 d&sigma;= ((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= &radic;(25- r2) and F= 25(rcos&theta;i+rsin&theta;j+ &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.)