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

MATH101 B April 2024

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[hide] 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 $V_{y}=2V_{x}$.

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[show] 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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[show] 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. 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 $a=1/2$.