Science:Math Exam Resources/Courses/MATH101/April 2007/Question 02 (c)

MATH101 April 2007

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Question 02 (c)
Full-Solution Problem. Justify your answers and show all your work. Simplification of answers is not required. Set up, but do not evaluate, a definite integral for the volume of the solid obtained by rotating the region between $y=0$ and $y=\sin x$ for $0\leq x\leq \pi$ about the line $y=1$.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
It might help to draw a picture.

Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. A picture is included below. Imagine yourself standing on the x-axis at x=-1 looking towards the rotated object. This top down view (the circles in the picture) show that what we are doing is taking a cylinder of radius 1 and height $\pi$, and subtracting from it the shape that we get from removing the inner shape of radius $(1-\sin(x))^{2}$. This gives the volume to be $V=\int _{0}^{\pi }\pi (1^{2}-(1-\sin(x))^{2})\,dx=\pi \int _{0}^{\pi }(2\sin(x)-\sin ^{2}(x))\,dx$ completing the question.

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