Finding volume of axis of revolution

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SUMMARY

The discussion centers on solving the integral for the volume of a solid of revolution, specifically the expression 2π∫₀⁶ y√(25 - (y - 1)²) dy + 5π. The user attempted to solve it using a substitution u = y - 1 and broke the integral into two parts, ultimately arriving at an approximate answer of 25π² + 499.89π/3. However, the expected exact answer is 25π² + 500π/3. The discrepancy arises from the evaluation of the second integral, which involves trigonometric substitution.

PREREQUISITES
  • Understanding of integral calculus, specifically volume of revolution
  • Familiarity with trigonometric substitutions in integrals
  • Knowledge of the properties of definite integrals
  • Experience with substitution methods in integration
NEXT STEPS
  • Review the method of trigonometric substitution in integrals
  • Study the evaluation of definite integrals involving square roots
  • Practice solving volume of revolution problems using different axes
  • Learn about the application of the Fundamental Theorem of Calculus in solving integrals
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Students studying calculus, particularly those focused on integral applications in volume calculations, and educators looking for examples of solving complex integrals.

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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