Science:Math Exam Resources/Courses/MATH101 A/April 2024/Question 19

MATH101 A April 2024

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Question 19
For a constant $a>0,$ consider the plane region lying above the curve $y=a^{2}x^{2}$ and below the curve $y={\sqrt {ax}}$. Rotating this region around the x-axis produces a solid: let $V_{x}$ denote the resulting volume. Similarly, let $V_{y}$ denote the volume produced by rotating the same region around the $y$-axis. Find the value of a for which $V_{y}=2V_{x}$.

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
Calculate $V_{x}$ and $V_{y}$ for a general $a$, using either washers or cylindrical shells. Then determine what $a$ must be in order for us to have $V_{y}=2V_{x}$.

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Solution

Washer (or annulus) cross-section of a volume of revolution Let us use the method of washers. (We could also use the cylindrical shells.) The curves intersect at $(0,0)$ and $(1/a,1)$. To find the volume of the solid formed by rotating the region around the $x$-axis, think of this solid as a sum of cross-sections. For each $x$ from $0$ to $1/a$, as shown in the figure to the right, the cross section at $x$ is a ring with outer radius $R={\sqrt {ax}}$ and inner radius $r=a^{2}x^{2}$. Thus, the area of this cross section is $\pi (R^{2}-r^{2})=\pi (ax-a^{4}x^{4})$. Each cross section should be thought of as having a "thickness" of $\mathrm {d} x$. Therefore, adding up the cross sections to find the volume, we get $${\begin{aligned}V_{x}=\int _{0}^{1/a}\pi \left(ax-a^{4}x^{4}\right)\;\mathrm {d} x=\pi \left.\left[{\frac {ax^{2}}{2}}-{\frac {a^{4}x^{5}}{5}}\right]\right|_{0}^{1/a}=\pi \left({\frac {a^{-1}}{2}}-{\frac {a^{-1}}{5}}\right)={\frac {3\pi }{10a}}.\end{aligned}}$$ For the solid formed by rotating the region around the $y$-axis, we must first rewrite the equations of the curves so that they give $x$ in terms of $y$, instead of $y$ in terms of $x$. The curve $y=a^{2}x^{2}$ can also be written as $x={\sqrt {y}}/a$, and the curve $y={\sqrt {ax}}$ can be written as $x=y^{2}/a$. We omit the figure in this case but encourage you to draw your own. You will see that for each $y\in [0,1]$, the cross section of this new solid at $y$ is a ring with outer radius $R={\sqrt {y}}/a$ and inner radius $r=y^{2}/a$. This cross section has an area of $\pi (R^{2}-r^{2})=\pi \left(y/a^{2}-y^{4}/a^{2}\right)$, and a "thickness" of $\mathrm {d} y$. By adding up all these cross sections, the volume is $${\begin{aligned}V_{y}=\int _{0}^{1}\pi \left({\frac {y}{a^{2}}}-{\frac {y^{4}}{a^{2}}}\right)\;\mathrm {d} y=\pi \left.\left[{\frac {y^{2}}{2a^{2}}}-{\frac {y^{5}}{5a^{2}}}\right]\right|_{0}^{1}={\frac {\pi }{a^{2}}}\left({\frac {1}{2}}-{\frac {1}{5}}\right)={\frac {3\pi }{10a^{2}}}.\end{aligned}}$$ Finally, let us find which $a$ makes it so that $V_{y}=2V_{x}$. We would like $a$ to satisfy ${\frac {3\pi }{10a^{2}}}=2\cdot {\frac {3\pi }{10a^{2}}}$. The solution is $a=1/2$.