We go through the usual routine to find the critical point, finding f_x = 3e^y - 3x² and f_y = 3xe^y - 3e^3y. Setting f_x equal to 0 and rearranging a little produces x² = e^y, and substituting this into the equation we get from setting f_y equal to zero produces 3x(x²) - 3(x²)³ = 0, or x³ = x⁶. The real roots of this equation are x = 0 and x = 1, but we discard x = 0 because there's no corresponding value for y that satisfies (0)² = e^y. Thus the only critical point we find is (1,0). We work out f_xx = -6x, f_yy = 3xe^y - 9e^3y, and f_xy = 3e^y. Then D(1,0) = [-6(1)][3(1)e⁽⁰⁾ - 9e³⁽⁰⁾] - [3e⁽⁰⁾]² = 27, which is positive, indicating an extremum. Then since f_xx(1,0) = -6 is negative, it must be a local maximum.

A pretty good view can be had from the Mathematica command Plot3D[3x E^y - x^3 - E^(3y), {x,-1.5,1.25},{y,-2.5,.5},Boxed->False,PlotRange->{-2,2.5},AxesLabel->{"x","y","z"},BoxRatios->{1,1,1},ViewPoint->{2,6,3}]. Our local max is toward the near left corner, but the far right corner rises higher yet.

The final issue is how the graph manages to accomplish this turnaround without having a local min, or at least a saddle point, somewhere in between. The answer has to do with the value of x = 0 that we discarded earlier. The graph has something like a "gully" between where x = 0 and where x = -1, where the function gets lower and lower as y increases but never becomes an actual critical point, yet does allow the function to turn around without having both partial derivatives equal zero.

 
2. Set up an iterated integral and use it to compute the volume of a rectangular box with sides of lengths a, b, and c.

If we situate the box with one vertex at the origin and extending into the first octant, so that its vertices are at (a,0,0), (0,b,0), (0,0,c), (a,b,0), (a,0,c), (0,b,c), and (a,b,c), then the integral looks like , which works out to abc.
 
 
3. Set up an iterated integral and use it to compute the volume of the solid with plane faces and vertices at (0, 0, 0), (a, 0, 0), (0, b, 0), and (0, 0, c).

We need a formula for the top surface of this solid, which is the plane passing through (a,0,0), (0,b,0), and (0,0,c). There are easier and harder ways to arrive at this, but probably quickest is to use the equation z = mx + ny + c for a plane with z-intercept c, and think about the slopes of our plane in the xz and yz planes. This gives us z = (-c/a)x + (-c/b)y + c.

We then need to determine our limits of integration in the xy plane, which involve the lines x = 0, y = 0, and the line where our top plane intersects the xy plane. Setting z = 0 in our top plane and rearranging gives us y = (-b/a)x + b. Then our integral is , which works out to abc/6.
 
 
4. Consider the truncated paraboloid bounded by the surfaces z = x² + y² and z = a, for some positive constant a. Express its volume as an iterated integral, and find its volume.
 
First determine the region on which to integrate by setting the two functions equal to each other, producing a = x² + y², a circle centered at the origin with radius .
 
This suggests we use cylindrical coordinates, so our paraboloid becomes z = r², while our top function remains z = a. Then our integral is .
 
 
5. Compare the volume of the truncated paraboloid from problem 4 to the volumes of the cone inscribed in it and the cylinder circumscribed around it.
 
You can either use calculus or use the old-fashioned formulas for volumes of cones and cylinders, but in either case you need to have a radius of  and height of a.
 
The integrals look like 
 
Looking at these, then, we see that the whole cylinder is twice the paraboloid and three times the cone, or V_cylinder = 2V_paraboloid = 3V_cone.
 
 
6. Looking at a graph of the function , we can see an infinite number of "ripples" (think of each "ripple" as the solid region above the xy plane) spreading outward from the origin. Find a formula for the volume of the n^th ripple, and show why it works.
 
Looking at the graph shows that the first ripple extends from a radius of 0 to a radius of pi, the second from 2pi to 3pi, and so forth, or in general the n^th ripple from 2(n-1)pi to 2(n-1)pi + pi.
 
The integral for the volume of the n^th ripple is then . Integrating produces something very messy like , but a little thought about the periodic properties of the trig functions lets us get rid of most of those "2n pi" terms and have just , which simplifies to .