Actually, if ojones had not suggesed spherical coordinates, I would have been inclined to use cylindrical coordinates for the first problem and certainly for this one. In cylindrical coordinates, the equation of the sphere is \(\displaystyle r^2+ z^2= 1\) and of the cone, z= r. Those intersect where \(\displaystyle r^2+ r^2= 2r^2= 1\) so at \(\displaystyle z= \frac{1}{\sqrt{2}\). At that z, the equation of the sphere becomes \(\displaystyle r^2+ 1/2= 1\) or \(\displaystyle r^2= 1/2\), \(\displaystyle r= \frac{1}{\sqrt{2}}\).

The volume under the sphere, above the plane of intersection of the cone and sphere, is \(\displaystyle \int_{\theta= 0}^{2\pi}\int_{r= 0}^{\frac{1}{\sqrt{2}}} (\sqrt{1- r^2}- \frac{1}{\sqrt{2}} r sin(\theta) drd\theta\).

I would like to add that in computations of volume bounded by a sphere/cylinder/cone/etc I first attempt the problem using cylindrical co-ordinates and I only use spherical co-ordinates if the form that I end up with is unrecognizable.

I do this for a few reasons: the first of which is cylindrical co-ordinates is simpler in that it only has 2 bounds, theta and radius, of course we have a dz integral but this is very easily computed that we don't even need to set it up. As opposed to phi, theta and P that spherical co-ordinates introduce. Also, spherical co-ordinates involve a very clumsy dV (at least in my opinion). The second thing is that cyldrincal co-ordinates are easier to visualize and interpret and most problems involving spheres and such are easier to do with cylindrical co-ordinates!

As for generally finding the volume bounded by a sphere and some shape (a cone, a plane, a cylinder, etc) we are best served by a diagram. Of course, most of us suck as graphing 3D objects (I am horrible) but we should be able to identify rudementary shapes such as a sphere, cylinder, cone, plane, parabola, etc.

For example,

Say I want the volume bounded by

\(\displaystyle z = 4 - x^2 - y^2 \) and \(\displaystyle z = 5+ x^2 + y^2 \)

Do I know how to graph that? Absolutely not, but I do know that they both involve parabolas in the XY plain (note the combination of \(\displaystyle x^2 + y^2 \) ) and since one is negative and one is posative, they must be parabolas interesecting eachother. We can then see that our upper z bound will be

\(\displaystyle z = 5+ x^2 + y^2 \)

And this is all we really need to do. Of course, this type of analysis isn't good for odd looking shapes and for those I go back to the definition of the triple integral

\(\displaystyle \int_a^b dx \int_{g_1 (x) }^{g_2 (x) } dy \int_{h_1 (x,y) }^{ h_2 (x,y) } dz \)

We simply fit the above form in whatever way we see fit by letting x,y equal 0 at specific times and find what z function is greater on the interval. Naturally this will only work for 95% of the cases because some are actually cancelled out by symetry and if you can't draw the object you are SOL in this case. But that is a rarity and we can't really be expected to see that, because I know professors who have a hard time computing those kinds of cases and it's unlikely they would expect students to know. Unless they provide the graph to begin with!