Triple integrals?

Question

set up but do not evaluate int[int[int[x^2+y^2+z^2dV]]] where Q is region bounded by z = root(16-x^-y^2) and the xy plane, and inside the planes y=0, x=0 for x>=0

Answer 1

The surface that forms the upper bound of Q has the following equation:
x^2 + y^2 + z^2 = 16
(This can be found by manipulating the equation you gave.)
This is the equation of a sphere with a radius of 4 and center at the origin. Since we are interested in the region above the XY plane (z>0), we are integration the given expression over the volume of a hemisphere.

Let's make the outer integral an integral in z. The limits for this integral are 0 to 4 (i.e., from the XY plane, where z is 0, to the top of the sphere, where z = 4 (and x and y are both 0).

Having defined the limits for z, we need to find limits for x and y in terms of z. So we ask ourselves, "For a given value of z (between 0 and 4), what are the limits of x and y?"

We could start with either one, but let's say we start with y. For a given value of z, what are the maximum and minimum values of y? Rewriting the above equation, we have:
y^2 = 16 - z^2 - x^2.
Now for a given z, the quantity (16 - z^2) will be a constant. x can vary, but we are interested in determining the largest and smallest values for y. These will occur when x is 0; when x is 0,
y equals the positive or negative square root of (16 - z^2), so the limits for y are sqrt(16-z^2) and -sqrt(16-z^2).

Finally, what are the limits on x, for a given value of z and a given value of y?
Again rewriting the equation, we have:
x^2 = 16 - z^2 - y^2
Treating both y and z as constants, we can immediately write:
x = sqrt(16-z^2-y^2) and -sqrt(16-z^2-y^2). These are the two values of x that are ON THE SURFACE of the sphere when y and z have the chosen values. If we CONNECT those two points, we are constructing a line segment THROUGH the sphere.

We then integrate with respect to x over this line segment. Then we integrate the resulting expression with respect to y over the limits for y (as determined above). At this point we have integrated over a disc (i.e., the interior of a circle), which represents the intersecion of the sphere with the plane that represents a particular value of z. Finally, we integrate the result of that integration once more, this time with respect to z over the range from 0 to 4. This represents an integration covering all the discs from the XY plane (z = 0) to the top point of the sphere (where z = 4). So we have now covered the entire interior of the hemisphere.

Trying to write this out without the benefit of a mathematical typeface, it looks something like this:

int(z=0 to 4) int(y= -sqrt(16-z^2) to sqrt(16-z^2)) int x= -sqrt(16-z^2-y^2) to sqrt(16-z^2-y^2))[x^2+y^2+z^2] dx dy dz

Hope this all makes sense to you.
Good luck.