Science:Math Exam Resources/Courses/MATH101/April 2010/Question 02 (a)

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MATH101 April 2010

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Question 02 (a)
Full-Solution Problem. Justify your answer and show all your work. Simplification of the answer is not required. The bounded region that lies between the $x$-axis and the curve $y=1-x^{2}$ is revolved about the line $y=-2.$ Find the volume of the resulting solid of revolution.

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
Sketch a diagram of the situation.

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Solution 1

We see the function and a sketch of the volume in the following figure. We use the shell method $V=2\pi \int h(z)z\,dz$ to find the volume. Therefore we need to determine what is the height $h(z)$ and what is the integral variable $z$. On the next figure we see, that the height is the inverse of the function $y=f(x)=1-x^{2}.$ And for this height the integral variable $z$ is the distance from $y$ to the rotating axis, which is $y=-2.$ So, we find z = y + 2 and for the height we calculate $f(x)=y=1-x^{2}\$ $x^{2}=1-y\$ $f^{-1}(y)=x={\sqrt {1-y}}.\$ For convenience, we calculate half of the volume and drop the left half. Then we can take $\displaystyle h(z)={\sqrt {1-y}}$ for z = y + 2. The boundaries of the integral must be the left and right edge of the interval for $y=[0,1]$. Now we need to calculate the integral ${\begin{aligned}{\frac {1}{2}}V&=2\pi \int h(z)z\,dz\\&=2\pi \int _{0}^{1}(y+2){\sqrt {1-y}}\,dy\\&=2\pi \overbrace {\int _{0}^{1}\underbrace {y} _{=u}\underbrace {\sqrt {1-y}} _{=dv}\,dy} ^{\text{use integration by parts}}+2\pi \int _{0}^{1}2{\sqrt {1-y}}\,dy\\&=2\pi \left[y(1-y)^{\frac {3}{2}}(-{\frac {2}{3}})\right]_{0}^{1}+2\pi {\frac {2}{3}}\int _{0}^{1}(1-y)^{\frac {3}{2}}\,dy+2\pi 2\left[(1-y)^{\frac {3}{2}}\right]_{0}^{1}\\&=0+2\pi {\frac {2}{3}}\left[(1-y)^{\frac {5}{2}}(-{\frac {2}{5}})\right]_{0}^{1}-0+2\pi {\frac {4}{3}}\\&=2\pi {\frac {4}{15}}+2\pi {\frac {4}{3}}\\&={\frac {16\pi }{5}}\end{aligned}}$ So, we get for the volume $V={\frac {32\pi }{5}}.$

Solution 2

We see the function and a sketch of the volume in the following figure. We use the washers method $V=\pi \int (R_{O}(x)^{2}-R_{I}(x)^{2})\,dx$ to find the volume, where RO is the outer radius of the washer and RI is inner radius. From the diagram above, we see that $R_{O}(x)=1-x^{2}+2=3-x^{2}$, and $R_{I}(x)=2$. We are integrating along x, so the boundaries of the integral must be $x=[-1,1]$. Now we need to calculate the integral ${\begin{aligned}V&=\pi \int (R_{O}(x)^{2}-R_{I}(x)^{2})\,dx\\&=\pi \int _{-1}^{1}((3-x^{2})^{2}-4)\,dx\\&=\pi \int _{-1}^{1}(9-6x^{2}+x^{4}-4)\,dx\\&=\pi \int _{-1}^{1}(5-6x^{2}+x^{4})\,dx\\&=\pi \left[5x-2x^{3}+{\frac {1}{5}}x^{5}\right]_{-1}^{1}\\&=\pi \left[\left(5-2+{\frac {1}{5}}\right)-\left(5(-1)-2(-1)^{3}+{\frac {1}{5}}(-1)^{5}\right)\right]\\&=\pi \left(10-4+{\frac {2}{5}}\right)\\&={\frac {32\pi }{5}}\end{aligned}}$

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