Disk/Washer Problem

[dBanji]

Hi everyone, how do I find the volume of these curves bounded by y = 7 − x, y = x, and x = 0 rotated around the y-axis. The question asks to use a disk method. I have used both the graphing calculator (after programming for disk/washer/shell) as well as working it out manually and kept getting the same result of 343pi/4, and 343pi/3 but none is correct. The second one is similar y=1/x; x = 2, and y = 4 (using disk method rotated around x-axis; I got 7pi/2 several times but wrong) Any help, please.

 

[Deleted member 4993]

Hi everyone, how do I find the volume of these curves bounded by y = 7 − x, y = x, and x = 0 rotated around the y-axis. The question asks to use a disk method. I have used both the graphing calculator (after programming for disk/washer/shell) as well as working it out manually and kept getting the same result of 343pi/4, and 343pi/3 but none is correct. The second one is similar y=1/x; x = 2, and y = 4 (using disk method rotated around x-axis; I got 7pi/2 several times but wrong) Any help, please.

Please share your "manual" work - so that we can catch any mistake (if there is any!).

Please show us what you have tried.

 

[dBanji]

Please share your "manual" work - so that we can catch any mistake (if there is any!).

Please show us what you have tried.

Please find attached one of the attempts. The work was sketchy and not well-scripted since I was only solving it for myself. I have been able to find one that is a bit presentable but the other one will not make sense if I post it now. When I settle down, I can work it again but maybe help with one will give me insight into solving the other.

thank you. And I read the instructions before posting.

 

[skeeter]

Both problems require use of the washer method (disk method with a "hole")

[MATH]V = \pi \int_0^{7/2} (7-y)^2 - y^2 \, dy[/MATH]
[MATH]V = \pi \int_{1/4}^2 4^2 - \left(\frac{1}{x}\right)^2 \, dx[/MATH]

 

[Steven G]

As skeeter is pointing out, the area you are revolving is from the curve y=1/x to the x-axis. This is exactly the area you do not want! If you revolved the area from y=4 to the x-axis then you get the volume you want and more. You need to subtract the more! This more is exactly the volume you obtain from revolving the are between y=1/x and the x-axis (y=0). You do not need to use integrals to calculate the volume when you revolve the area between y=4 and y=0. Why is that? What type of shape would you get?

 

[dBanji]

Both problems require use of the washer method (disk method with a "hole")

[MATH]V = \pi \int_0^{7/2} (7-y)^2 - y^2 \, dy[/MATH]
[MATH]V = \pi \int_{1/4}^2 4^2 - \left(\frac{1}{x}\right)^2 \, dx[/MATH]

Thank you very much. The second one worked, I think I ignored the y=4 since it stands without x. Thank you. The second one however results in the same result I have always gotten (343pi/4). Done it more than 5 times again the same result.

 

[skeeter]

[MATH]\pi \bigg[16x + \frac{1}{x} \bigg]_{1/4}^2 = \pi \bigg[\left(32+\frac{1}{2}\right) - (4 + 4)\bigg] = \frac{49\pi}{2}[/MATH]

 

[nasi112]

Using Disk Method for the first one.

[MATH]\pi\int_{0}^{3.5} y^2 \ dy + \pi\int_{3.5}^{7} (7 - y)^2 \ dy = \frac{343\pi}{12}[/MATH]
Or you can shorten the integral because it is symmetric.

[MATH]2\pi\int_{0}^{3.5} y^2 \ dy = \frac{343\pi}{12}[/MATH]
And you can also use the volume of the cone to make sure your answer is correct.

[MATH]V = \frac{1}{3}\pi(3.5)^2 \cdot 3.5[/MATH]
You have two cones, then

[MATH]2V = \frac{2}{3}\pi(3.5)^2 \cdot 3.5 = \frac{343\pi}{12}[/MATH]

 

[skeeter]

Using Disk Method for the first one.

[MATH]\pi\int_{0}^{3.5} y^2 \ dy + \pi\int_{3.5}^{7} (7 - y)^2 \ dy = \frac{343\pi}{12}[/MATH]
Or you can shorten the integral because it is symmetric.

[MATH]2\pi\int_{0}^{3.5} y^2 \ dy = \frac{343\pi}{12}[/MATH]
And you can also use the volume of the cone to make sure your answer is correct.

[MATH]V = \frac{1}{3}\pi(3.5)^2 \cdot 3.5[/MATH]
You have two cones, then

[MATH]2V = \frac{2}{3}\pi(3.5)^2 \cdot 3.5 = \frac{343\pi}{12}[/MATH]

I disagree with your result with the volume of the region rotated about the y-axis.

[MATH]V = \pi \int_0^{7/2} (7-y)^2 - y^2 \, dy[/MATH]
[MATH]V = \pi \int_0^{7/2} 49-14y \, dy[/MATH]
[MATH]\pi \bigg[49y - 7y^2 \bigg]_0^{7/2} = \pi \bigg[\frac{343}{2} - \frac{343}{4}\bigg] = \frac{343\pi}{4}[/MATH]

 

[nasi112]

I disagree with your result with the volume of the region rotated about the y-axis.

[MATH]V = \pi \int_0^{7/2} (7-y)^2 - y^2 \, dy[/MATH]
[MATH]V = \pi \int_0^{7/2} 49-14y \, dy[/MATH]
[MATH]\pi \bigg[49y - 7y^2 \bigg]_0^{7/2} = \pi \bigg[\frac{343}{2} - \frac{343}{4}\bigg] = \frac{343\pi}{4}[/MATH]

Skeeter. I have showed two methods to find the volume, and now I will show you the third method which is the washer method.

[MATH]2\pi\int_{0}^{3.5} x(7 - x - x) \ dx = 2\pi\int_{0}^{3.5} 7x - 2x^2 \ dx = \frac{343\pi}{12}[/MATH]

 

[dBanji]

Thank you both for helping out on this. I got 343pi/4 several times but was wrong. I cannot confirm this last answer of 343pi/12 as I have run out of attempts. It is good to know what the exact result is because something similar might show up in my exam (that was a homework problem).

I really appreciate you guys creating the time to look into this. It was so weird and I was surprised why the result was wrong. I might have to check with the professor to be sure there isn't anything wrong with the answer they provided.

 

[nasi112]

I understand why you got a wrong answer and why skeeter is also getting a wrong answer.



Look at the picture. you are rotating the green region about the y-axis. It is the wrong region.



This is the right region. The purple one.

 

[skeeter]

It happens ...