Math 223 Homework 5 Solutions

1. We can think of the volume of the capsule as being the volume of the solid region between the end caps, that is bounded in the xy plane by the cicle . This circle is the cross-sectional area of the capsule, if we orient the capsule so that the z axis lies along the length and the origin is at the center of the capsule. The endcaps can be thought of as surfaces in the form z = f(x, y) where the function f is the equation of the upper portion of a sphere of radius 2mm, centered at the point (0,0,5) for the upper cap. The lower cap is the bottom portion of a sphere of radius 2mm, centered at the point (0,0,-5). Thus, we have,

   

   on the top of the capsule, and

   

   on the bottom of the capsule. Thus, we can find the volume between these two surfaces by performing an interated integral over the circle of the difference of these two surfaces. In other words,

   

   It is much easier to set this up in polar coordinates, and to simplify the function that we are integrating. Thus, we have

   

2. To set up a three dimensional volume integral, we can simply use the equations above for and as limits of integration in the z direction, keeping the limits in the x and y directions the same as before. Then, we use cylindrical coordinates to get

   

   which means that the capsule has a total volume of cubic mm.

3. To calculate the mass of an object in three dimensions, we integrate the density of the object over the volume. Thus, the mass of the capsule is

   

   If we let we see that with u = 7 when r = 0 and u = 5 when r = 2. Thus, the mass integral becomes

   

   Thus, each capsule contains approximately 1.3357 mg of DEATH.

4. If an AIDS patient needs to take two capsules every twelve hours, then he or she must take 4 capsules per day. Thus, a patient would have to take of DEATH each day. Since the fatal dosage is 5 mg each day, this drug is not a viable drug.