The region between the graphs of $ y=x^2$ and $y=3x$ is rotated around the line $x=3$

What is the volume of the resulting solid?

I drew the picture, and I saw that I should be using the washer method. Since the region was being flipped at $x = 3$, I would have to make sure I am subtracting the two functions from 3 and then subtract those two, but I could not seem to get the right answer.


Using washers, the interval of integration will be along the $y$-axis from $y = 0$ to $y = 9$. The inner radius as a function of the $y$-value will be $r(y) = 3 - \sqrt{y}$, and the outer radius will be $R(y) = 3 - y/3$; thus the differential volume of a washer with thickness $dy$ is $$dV = \pi(R(y)^2 - r(y)^2) \, dy = \pi((3 - y/3)^2 - (3-\sqrt{y})^2) \, dy,$$ and the total volume is given by the integral $$V = \pi \int_{y=0}^9 (3 - y/3)^2 - (3-\sqrt{y})^2 \, dy.$$

Using cylindrical shells, the interval of integration will be along the $x$-axis from $x = 0$ to $x = 3$. The height of the shell is given by the difference of the functions $y = x^2$ and $y = 3x$, so $h(x) = 3x - x^2$ (since $3x \ge x^2$ on $x \in [0,3]$). The circumference of a representative shell is simply $2\pi(3-x)$, thus its differential volume is $$dV = 2\pi(3-x)(3x-x^2) \, dx,$$ and the total volume is $$V = \int_{x=0}^3 2\pi(3-x)(3x-x^2) \, dx.$$ Both calculations give the same result.


Volume by washers can be used in this case if you stack your washers parallel to the $x$-axis, centered at $x=3$. Identify the outer radius $R$ (as a function of $y$, since every washer is determined by how high it's sitting in the stack) and the inner radius (again, as a function of $y$); the volume of each thin washer is $\pi(R^2 - r^2)\,\Delta y$ and the total volume is $\pi\int_0^9(R^2 - r^2)\,dy$.

  • $\begingroup$ The outer radius would be $(y/3) - 3$ and the inner would be $sqrt(y)-3$, correct? Could you explain why the upper limit of the integral is 9? I was getting 3. Thank you for the help! $\endgroup$ – Sherbs Jan 7 '15 at 16:42
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    $\begingroup$ @user138002 You're stacking parallel to the $y$-axis, from $y=0$ to ... at what $y$ value do the curves intersect? And the inner radius is the distance from $x=3$ to the curve $x = \sqrt y$, so $r = 3-\sqrt y$; similar for $R$. Almost there! $\endgroup$ – Jon Jan 7 '15 at 16:57

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