The question I am trying to solve is as follows:
Use the Midpoint Rule with $n = 5$ to estimate the volume obtained by rotating about the $y$-axis the region under the curve $y = \sqrt{1+x^3}$, $0\le x\le1$.
The answer's supposed to be: $3.68$.
I just want to know the exact steps to solve the question; I couldn't get the same answer.
EDIT: Here is how I did it:
- A cross-section from the top would have a hole, so it's a washer.
- The lower y bound is 0 & the upper y bound is sqrt(2).
- Then, I found the y midpoints (midpoint rule).
- The first 3 intervals/slices have r=1, so their combined volume is (3)(pi)sqrt(2)/5.
- The 4th interval/slice has outer r=1 & inner r=7sqrt(2)/5, so its volume is (pi)[sqrt(2)/5]{1-[7sqrt(2)/5]^2}.
- For the 5th interval/slice it has outer r=1 & inner r=9sqrt(2)/5, so its volume is (pi)[sqrt(2)/5]{1-[9sqrt(2)/5]^2}.
- When I calculated it, I got abt. 2.133
Thanks in advance,
Your inner radius should be $r=\sqrt[3]{y^2-1}$ when it is greater than zero. The five midpoints are at $(0.1,0.3,0.5,0.7,0.9)\sqrt 2.$ I find only the last has a hole in the middle of radius $\sqrt[3]{(0.9\sqrt 2)^2-1}\approx 0.8527$.