Finding volume of axis of revolution

In summary: 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?
  • #1
Sidthewall
33
0

Homework Statement



[tex]
2\pi \int_0^6 y\sqrt{25- (y- 1)^2}dy + 5\pi
[/tex]
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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  • #2
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.
 
  • #3
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
 
  • #4
Sidthewall said:

Homework Statement



[tex]
2\pi \int_0^6 y\sqrt{25- (y- 1)^2}dy + 5\pi
[/tex]
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
 

FAQ: Finding volume of axis of revolution

1. What is the axis of revolution?

The axis of revolution is an imaginary line around which an object is rotated to create a three-dimensional shape.

2. How do you find the volume of an axis of revolution?

To find the volume of an axis of revolution, you can use the formula V = π∫(R(x))^2dx, where R(x) is the radius of the cross-section at a given point along the axis of revolution.

3. What are some common objects that can be measured using the axis of revolution method?

Some common objects that can be measured using the axis of revolution method include spheres, cones, cylinders, and tori (donuts).

4. What is the difference between using the axis of revolution method and other methods for finding volume?

The axis of revolution method is specifically used for finding the volume of objects that can be created through rotation, whereas other methods may be used for more general shapes.

5. Are there any limitations to using the axis of revolution method?

Yes, the axis of revolution method can only be used for objects that have rotational symmetry, meaning they look the same after being rotated around an axis. It also only works for objects with smooth, continuous surfaces.

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