Washer method "solid of revolution" problem

calc67x

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Feb 6, 2018
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Problem with setting up Washer/Disk method for determining solid of revolution.

Need to find the volume of the solid of revolution:


Find the volume of the solid of revolution generated by revolving region about the line y=7

y=7sqrtx, y=7 and x=0

I set up the problem like this:

a=0 b=1 pi[7^2 - (7sqrtx)^2 dx] using the washer method

The answer to this problem is 49pi/6 in my book.
I got 49pi/2.

How am I setting up the problem wrong? These are difficult problems, and getting an answer does not help with the learning.
 

HallsofIvy

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Cross sections are "disks", not "washers", because there is no hole- the figure has y= 7 as one boundary and rotation is around y= 7. A line segment, perpendicular to y= 7, ending at \(\displaystyle y= 7\sqrt{x}\) has length \(\displaystyle 7- 7\sqrt{x}= 7(1- \sqrt{x})\). So a disk at a given x will have radius \(\displaystyle 7- 7\sqrt{x}= 7(1- \sqrt{x})\) and area \(\displaystyle \pi(7- 7\sqrt{x})^2= 49\pi(1- \sqrt{x}))^2\) so a disk with thickness \(\displaystyle \Delta x\) has volume \(\displaystyle 49\pi(1- \sqrt{x})^2\Delta x\). The volume is approximately the sum of those, \(\displaystyle 49\pi\sum(1- \sqrt{x})^2\Delta x\). In the limit, as the thickness goes to 0 and the number of disks goes to infinity, is the integral \(\displaystyle 49\pi\int_0^7 (1- \sqrt{x})^2 dx\).
 
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Jomo

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Dec 30, 2014
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Problem with setting up Washer/Disk method for determining solid of revolution.

Need to find the volume of the solid of revolution:


Find the volume of the solid of revolution generated by revolving region about the line y=7

y=7sqrtx, y=7 and x=0

I set up the problem like this:

a=0 b=1 pi[7^2 - (7sqrtx)^2 dx] using the washer method

The answer to this problem is 49pi/6 in my book.
I got 49pi/2.

How am I setting up the problem wrong? These are difficult problems, and getting an answer does not help with the learning.
Professor Halls showed you how to do the problem with the disc method (because you used dx). Now if the problem said to use the washer method then here is how you would proceed.

a=o to b=7 (2pi*r*x)dy = a=o to b=7 (2pi*(7-y)(y2/49)dy

These are only difficult if you do not do lots of them. When they start all looking the same, then you are done studying. Until then, you need to practice more.
 

calc67x

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Joined
Feb 6, 2018
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Thanks!

Great, this is exactly what I was looking for!

Professor Halls showed you how to do the problem with the disc method (because you used dx). Now if the problem said to use the washer method then here is how you would proceed.

a=o to b=7 (2pi*r*x)dy = a=o to b=7 (2pi*(7-y)(y2/49)dy

These are only difficult if you do not do lots of them. When they start all looking the same, then you are done studying. Until then, you need to practice more.
 

Dr.Peterson

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Nov 12, 2017
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Great, this is exactly what I was looking for!
HallsofIvy showed you the disk/washer method, which involves circular or annular slices perpendicular to the axis. Jomo's method is actually the cylindrical shell method, where strips parallel to the axis are revolved around the axis. So that is not really what you were asking for.
 

Jomo

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Dec 30, 2014
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HallsofIvy showed you the disk/washer method, which involves circular or annular slices perpendicular to the axis. Jomo's method is actually the cylindrical shell method, where strips parallel to the axis are revolved around the axis. So that is not really what you were asking for.
Yes, of course that is the shell method. It's been so long that I forget the name. At least I did it correctly.
 
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