Help finding volume of solid using disk/washer method

[R.M.]

I am attempting to find the volume of a solid, but I found two conflicting answers, neither of which agree with mine.

Problem: "Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line.
y=[MATH]\bbox{\frac{1}{4}}[/MATH]x², x=2, y=0, about the y-axis"

My attempted solution:
Graphed the region, then sketched the resulting solid (kinda sorta).




Since I'm rotating about the y-axis, I need a horizontal cross-section of height [MATH]\Delta y[/MATH]. Rewriting the original function in terms of y gives x=[MATH]\pm2\sqrt {y}[/MATH].

R = 2 and r = 2[MATH]\sqrt{y}[/MATH]. Therefore area of cross-section is [MATH]\pi[/MATH](R²-r²) = [MATH]\pi[/MATH](4-4y).

[MATH]\pi\int_0^1 (4-4y) \,dy[/MATH] = [MATH]\pi[/MATH](4y-2y²[MATH]\Biggr|_{0}^{1}[/MATH] = 2[MATH]\pi[/MATH].

When I went to verify my solution, one source said the answer was [MATH]\frac{\pi}{6}[/MATH] and another source said the answer was 16[MATH]\pi[/MATH]. If I'm incorrect, I'm not sure where I made a mistake. Thank you for any advice/help.

 

[tkhunny]

One may wish to start with the Filled-In Right Circular Cylinder. [math]\pi\cdot 2^{2}\cdot 1 = 4\pi[/math]. This rather firmly rules out [math]16\pi[/math]

 

[Dr.Peterson]

When I went to verify my solution, one source said the answer was [MATH]\frac{\pi}{6}[/MATH] and another source said the answer was 16[MATH]\pi[/MATH]. If I'm incorrect, I'm not sure where I made a mistake. Thank you for any advice/help.

It would appear that your mistake was in selecting your sources! Are you sure they are solving the same problem? Are they reliable? What were they?

 

[MarkFL]

Using the washer method:

[MATH]dV=\pi(2^2-4y)\,dy=4\pi(1-y)\,dy[/MATH]
Hence:

[MATH]V=4\pi\int_0^1 1-y\,dy=4\pi\left[y-\frac{1}{2}y^2\right]_0^1=2\pi[/MATH]
Shell method:

[MATH]dV=2\pi x\cdot\frac{1}{4}x^2\,dx=\frac{\pi}{2}x^3\,dx[/MATH]
Hence:

[MATH]V=\frac{\pi}{2}\int_0^2 x^3\,dx=\frac{\pi}{8}\left[x^4\right]_0^2=2\pi[/MATH]
So, I do agree with your result.

 

[R.M.]

Thank you so much. I thought I was losing my mind! I have taught high school math for the last 18 years (mostly geometry, algebra 2, and most recently precalculus), and may be assigned to teach calculus as a CAPP (college credit) course soon. For the past year or so, I've been slowly working my way through a calculus book to review the concepts and make sure I can solve problems that I may assign. I have answers to odds, so this particular one I punched into google to check my answer. After re-reading the 2 "sources" I checked, I discovered why they didn't match my answer. One had the correct integration bounds, but wrote the integral in terms of x and dx; the other one had the integral in terms of y and dy, but took the integral over [-2, 2].

 

[R.M.]

One source was yahoo answers (face palm). Appears whoever wrote that used disks rather than washers. The other one was a bit more legitimate in my opinion: https://math.berkeley.edu/~buehler/1A-Fall2012/Worksheet31-Solutions.pdf

Solution is on page 4. Was puzzled why it didn't match mine, until I realized that the integral bounds are x-values, while the function being integrated is in terms of y.

 

[MarkFL]

I forgot to say:

Hello, and welcome to FMH!

You did a great job of presenting the problem, and showing your work, and answering posted questions. Model forum behavior.

 

[HallsofIvy]

Notice that you can also do this by calculating the volume of the "hole" and subtracting that from the volume of the cylinder you get by rotating 0< x< 2, 0< y< 1 around the y-axis. That cylinder has radius 2 and height 1 so its volume is \(\displaystyle 4\pi\). The "hole" is the parabola \(\displaystyle y= x^2/4\) with x= 0 to 2, y from \(\displaystyle y= x^2/4\) to 1. A "disk" at a given y will have radius \(\displaystyle x= 2y^{1/2}\) so area \(\displaystyle 4\pi y\) and, with thickness "dy", volume \(\displaystyle 4\pi y dy\). The total volume of the "hole" is \(\displaystyle 4\pi \int_0^1 ydy= 2\pi\). The volume of the original region is \(\displaystyle 4\pi- 2\pi= 2\pi\).