A student was asked to find the volume of the bowl-shaped figure generated when the curve y = x2 from x = 0 to x = 2 is revolved around the y-axis. She used the disk method and found the volume to be \displaystyle \int_{0}^{4}{\pi ydy}=8\pi . To check her work she used the method of cylindrical shells and found the same answer: \displaystyle\int_{0}^{2}{2\pi x\left( 4-{{x}^{2}} \right)dx}=8\pi .

The second part of the question asked for the value of x for which the bowl would be half full. So she first solved the equation \displaystyle \int_{0}^{k}{\pi ydy=\frac{1}{2}\cdot 8\pi }  and found k=2\sqrt{2}. This is a y-value so the corresponding x-value is \sqrt{2\sqrt{2}}\approx 1.682. She again checked her work by shells by solving \displaystyle \int_{0}^{k}{2\pi x\left( 4-{{x}^{2}} \right)dx=\frac{1}{2}\cdot 8\pi } and found that k=\sqrt{-2\left( \sqrt{2}-2 \right)}\approx 1.082. (This is the x-value.)

Both computations are correct. Can you explain to her why her answers are different? Use the comment box below to share your explanations. I will post mine in a week or so.


2 thoughts on “Challenge

  1. Pingback: Challenge Answer | Teaching Calculus

  2. Maybe I’m just oversimplifying this, but I think about picturing how the solid is being “filled up” by each method. For the disc method, we start from the “bottom” of the solid, whereas the shells method starts from the middle of the solid and works its way out to the edges. Therefore each method will reach halfway at a different point. If I were to choose one for applicability (is that a word?) I would use the disc method, since that models what is more likely to happen in an actual situation (thanks to gravity).


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