The Conundrum of Rotating Cakes

AP Calculus AB 2010 Free Response #4 is quite the conundrum, but cakes, once again, will analogically save the day.

The problem has to do with this:

But cake makes it better so:

Let "R" (or the cake) be the region in the first quadrant bounded by the graph of y = 2√(x), the horizontal line y = 6, and the y axis.

(a) Find the area of "R", or, in this case, the area of the cake.

The best way to think about this is to imagine the cake being a whole rectangle:

Now you can recognize that the area of the cake is the area of the whole cake minus the area that the y = 2√(x) cuts out. The area of the cake is 54 units through multiplication, but ,in order to do this the Calculus way, you'll need to subtract the area under the curve (or integral) of the y = 2√(x) from the area under the curve (or integral) of y = 6 from x = 0 to x = 9:

(b) Write, but do not evaluate, an integral expression that gives the volume of the cake generated when "R" is rotated about the horizontal line y = 7.

This is where it gets much more tricky. Remember that there are three ways to integrate solids generated by rotation: disk, washer, and the shell method. In this case, we will have to use the washer method because, like a washer, there will be a hole in the center of the solid. You can tell this because you're rotating it at y = 7 which is greater than the y = 6 that the graph extends to. Therefore, you know that there will be a hole in between y = 6 and y = 7.

The rotation would look something like this:

Rotate along y = 7, get cake.

Now, you must determine how to use the washer method. Remember that the format for it is like this:

It is possible to reason through and get this equation if you don't memorize it. Remember that the 'pi' is there because you're rotating something in a circle, the integral is the domain (or range in case of dy) of the thing you are rotating, the outer and inner radii are squared which is similar to squaring the radius in order to get the area of a circle, and you subtract the outer radius from the inner radius because of the hole that is left by the inner radius. 

Now all we have to do is determine our "outer" radius which is:

The reason why this is the outer is because it gives the equation for all of the area under y = 7, but not the 'hole' on the inside. The hole on the inside, when rotated, is just a cylinder with a radius of r = 1, so all the equation for the inside radius is simply:

Now all you have to do is plug into the equation, and you're done:

(c) Region R is the base of a cake. For each y, where 0 < y < 6, the cross section of the cake taken
perpendicular to the y-axis is a rectangle whose height is 3 times the length of its base in region R. Write,
but do not evaluate, an integral expression that gives the volume of the solid.

For this, we will refer back to this picture:

Now all you have to do is imagine that there are a large number of rectangles stacked on top of each other going from left to right on the cake portion of the graph. In order to calculate what the estimate of a volume these rectangles is, you must first find them in terms of 'dy' because they are perpendicular to the y axis. To do this, solve the original equation for 'x':

And since the problem tells you that the height of these rectangles is three times greater than the base, then:

Now, with the height and the base, you can find the area of ONE of the rectangles:

After getting the equation for ONE of the rectangles, the full rectangular estimate of the volume can be found by taking the integral of this equation:

The integral is taken from 0 to 6 because the base is perpendicular to the y axis and that base goes from 0 to 6. The integral is needed because it adds up all of the infinitely thin rectangles within the volume of the cake in order to get an estimate of the actual volume.

This problem was extremely 'thick'. If you need any clarification on the concepts needed to solve this problem or if you have any corrections or suggestions, do not hesitate to contact me at jonathan352@gmail.com.