Finding volume of axis of revolution

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Homework Help Overview

The discussion revolves around finding the volume of a solid of revolution using an integral involving a function of y. The integral presented is related to rotating a curve around an axis, and participants are examining the correctness of the integral setup and the calculations involved.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to solve the integral by substituting variables and breaking it into parts. Some participants question the validity of the integral setup and whether it accurately represents the problem. Others express uncertainty about specific calculations and results derived from the integral.

Discussion Status

Participants are actively discussing the integral and its components, with some providing insights into their calculations. There is a recognition of potential discrepancies in the results, and while some guidance has been offered regarding the calculations, there is no explicit consensus on the correctness of the approaches taken.

Contextual Notes

There is mention of imposed homework rules requiring the use of y for the integral, which may influence the approach taken by the original poster. Additionally, the complexity of the integral is noted, as it is described as a longer and harder method to find the volume.

Sidthewall
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Homework Statement



<br /> 2\pi \int_0^6 y\sqrt{25- (y- 1)^2}dy + 5\pi<br />
That's the integral i need solved

2. The attempt at a solution[/b]
so first i subbed u=y-1
took the 2 pi out of the integral
that got me 2 integrals u*sqrt(25-(u)^2) du + sqrt(25-(u)^2) du
the first integral = (2*pi*24^(3/2))/3

the second i used a trig sub and my final answer is
(25pi^2)/2 +
-25*asin(-.2)+sqrt(24)

The final answer is suppose to be 25pi^2 + 500pi/3

but i got25pi^2 + (-75*asin(-.2)+3*sqrt(24)+2*(24)^(3/2) -15)/3
which is approx
25pi^2 + 499.89pi/3

how do i get the exact answer
 
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I infer from the problem title that you are rotating some curve around some axis. Are you sure that your integral is the right one for your problem? I don't want to spend any time with the integral you have if I'm not sure it represents the problem you're working.
 
Yes I am 100% sure I am using the right integrel. The integral itself is odd and a longer and harder way to find the volume of this object. I would normally solve this in respect to x but the instrucotr wants it in respect to y.
And yeah I am 100% sure it is the right integrel
 
Sidthewall said:

Homework Statement



<br /> 2\pi \int_0^6 y\sqrt{25- (y- 1)^2}dy + 5\pi<br />
That's the integral i need solved

2. The attempt at a solution[/b]
so first i subbed u=y-1
took the 2 pi out of the integral
that got me 2 integrals u*sqrt(25-(u)^2) du + sqrt(25-(u)^2) du
the first integral = (2*pi*24^(3/2))/3
That looks fine.
Sidthewall said:
the second i used a trig sub and my final answer is
(25pi^2)/2 +
-25*asin(-.2)+sqrt(24)
I'm not sure about this part, especially sqrt(24). My answer for the 2nd integral, before being multiplied by 2pi and without adding 5pi was
25pi/4 - 25/2 * sin(2arcsin(-1/5)) - 25/2 * arcsin(-1/5).

I don't have time to check that before going to work, so I might have made a mistake.


Sidthewall said:
The final answer is suppose to be 25pi^2 + 500pi/3

but i got25pi^2 + (-75*asin(-.2)+3*sqrt(24)+2*(24)^(3/2) -15)/3
which is approx
25pi^2 + 499.89pi/3

how do i get the exact answer
 

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