Calculus IV Problem Set 2 Hints Spring 1999 Due 2/22/99


1. [Stewart, Problems Plus, p.830, #8] The plane x + y + z = 24 intersects the paraboloid z = x2 + y2 in an ellipse. Find the highest and lowest points on the ellipse.

Hints:

This can be handled very much like Example 5 in section 12.8.

Each of the equations given here can be viewed as one of the constraints.

The function to be optimized is a lot like the one in Example 5, but it's really just the height (the z coordinate) that we want to maximize, so our objective function is just f(x,y,z)=z.

2. [Inspired by Judith V. Grabiner's "'Some Disputes of Consequence': Maclaurin among the Molasses Barrels," from Social Studies of Science 28/1 (February 1998) 139-68] In 1735 the great British mathematician Colin Maclaurin "wrote a 94-page memoir to the Scottish Excise Commission, explaining how to gauge, with a single dip of a dipstick, the amount of molasses in the barrels in the Port of Glasgow." [p. 139] In this treatise he proved several surprising theorems to the general effect that the difference between the frustum of the solid produced by revolving a conic section around one of its axes and an approximating cylinder matching the radius at the midpoint of the frustum depends only on the height of the frustum. In particular, he proved that for a paraboloid of revolution the volume of the frustum is the same as the volume of the cylinder. Use a double integral to express the volume of a frustum of a paraboloid of revolution and show why this is true.

Hints:

Make sure you're computing the volume inside the paraboloid instead of what's underneath it.

Note it's asking about any paraboloid of revolution, not just z = x2 + y2, so there's a constant to be brought in somewhere.

The easiest way for us at this stage to think about frustums is as the difference between two paraboloids: The one up to a top height minus the one up to a bottom height.


3. Maclaurin also showed that the difference between a frustum of a right circular cone and the corresponding cylinder is one-fourth the volume of a similar cone, with the same height as the frustum and with diameter one-half the difference between the upper and lower diameters of the frustum. Use a double integral to express the volume of a frustum of a right circular cone and show why this is true.

Hints:

Everything from #2 still applies

My best advice is to handle each part separately: Find the volume of the frustum, find the volume of the approximating cylinder, and find the volume of the Cylinder that represents the difference.  Once you've got each individually, see if they fit together the way Maclaurin says.


4. Suppose the temperature in degrees Fahrenheit at a point (x,y) in a room which measures five meters (from 0 to 5) along the x axis by six meters (from -3 to 3) along the y axis is given by the function .

    a) Produce a graphic representation of the temperature in the room.

Hint: Temperature isn't something we intuitively associate with a surface.  A DensityPlot might be easier to interpret.
    b) What sort of physical situation might produce such a temperature distribution?
Hint: You're getting warmer....
    c) What is the average temperature in the room, to the nearest tenth of a degree?
Hint: Problems 35-36 in section 13.2 tell you how to compute an average volume.  Once you've got the integral set up, Mathematica will be happy to evaluate it for you.  Don't worry about exact values, they're impossible on this, that's why it only asks for an approximation.


5. a) Produce a graph of the surface x2/3 + y2/3 + z2/3 = 1.

Hints:
    Mathematica has trouble with those 2/3 exponents (I think it's considering imaginary roots, which we don't want to mess with).  To avoid problems, enter x2/3 as (x^2)^(1/3), and so forth.

    Keep in mind there's a top and a bottom, and you should show both together.
    PlotPoints, PlotRange, and BoxRatios are all good options to keep in mind here.

    b) Write an integral which would give the volume of the region inside the surface from part a.
Hints:
Think about the symmetry you see in the graph and use it.

As usual, the top view of the region's projection into the xy-plane is very useful for setting up your limits of integration.

    c) Compute the volume of the region from part b.
Hints:
This is by far the hardest problem here.  Be careful with it.

The substitution y = sin3(1-x2/3 )3/2 does wonders.

When you make the above substitution, go ahead and update your limits of integration to the new  variable.  I know I said in class this generally isn't worthwhile in multiple integrals, but this time it's a wonderful simplification.

The integral that's left still isn't easy at all, but it's doable.  Either use the half-angle identities or (the saner route) let Mathematica deal with it from here -- you've gotten it past the part where it was stuck.

Take the time to decide whether your answer is reasonable or not.  An answer of 762 is probably unreasonable for a region that lies well with a 2 by 2 by 2 cube, for instance.  Significantly better bounds can be had if you think about your graph carefully.