Science:Math Exam Resources/Courses/MATH101/April 2005/Question 02 (b)

MATH101 April 2005

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Question 02 (b)
Let $\displaystyle R$ be the finite region bounded above by the curve $\displaystyle y=4-x^{2}$ and bounded below by $\displaystyle y=2-x$. Express the volume of the solid obtained by rotating $\displaystyle R$ about the $x$-axis as a definite integral. You do not need to simplify or evaluate this integral.

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
Can you sketch a picture of this volume? How can this help you set your work?

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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. We would use disc integration to evaluate this volume. First, we need to determine the $x$ values that define the endpoints of the interval that we need to integrate over. The endpoints are where the curves intersect, so we solve $4-x^{2}=2-x$ or $x^{2}-x-2=0\quad \rightarrow \quad (x-2)(x+1)=0$ Thus, the endpoints of the interval $x=-1$ and $x=2$. Since we are using the disc method, we recall that the volume of a single disc with radius $r$ and thickness $\Delta x$ is $\pi r^{2}\Delta x$. Since we are rotating each curve about the $x$-axis, we can treat the value of the function as the height of each disc. By taking a sample point in the interval $(-1,2)$, we can see that $4-x^{2}>2-x$ in the domain of integration so we can get the volume of the defined solid by taking the volume of the solid with radius $4-x^{2}$ and subtracting the volume with radius $2-x$. ${\begin{aligned}V&=\int _{-1}^{2}\pi (4-x^{2})^{2}\,dx-\int _{-1}^{2}\pi (2-x)^{2}\,dx\\&=\pi \int _{-1}^{2}(4-x^{2})^{2}-(2-x)^{2}\,dx\end{aligned}}$ (Note: If you wish to evaluate the integral, the volume of the solid obtained by rotating the region $R$ about the $x$-axis is $V=108\pi /5$.)

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