Finding volume of axis of revolution

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The discussion revolves around solving the integral 2π ∫₀⁶ y√(25 - (y - 1)²) dy + 5π to find the volume of a solid of revolution. The user attempted to solve it by substituting u = y - 1 and breaking the integral into two parts, but their final result deviated from the expected answer of 25π² + 500π/3. There is uncertainty regarding the correctness of their calculations, particularly in the second integral, which involved trigonometric substitution. Despite confirming the integral setup, the user is seeking guidance on achieving the exact answer. The conversation highlights the complexity of integrating with respect to y and the challenges in verifying the calculations.
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
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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