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Mutaja
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Homework Statement
A function is given at y = [itex]\sqrt{x-x^2}[/itex]
Find the volume of the solid of the revolution around the x axis. Not sure if this is the correct translation, but find the volume when you revolve the function around the x axis.
The object in question is a sphere. What does the radius of the sphere has to be?
The Attempt at a Solution
I am slightly familiar with the disc method in this case, but I am not familiar with the notes I've written from my lectures. This is what I've done so far.
The function goes from 0-1, and when revolved it forms a sphere.
Radius: [itex]\sqrt{x-x^2}[/itex]
Area: ∏*##r^2## = ∏([itex]\sqrt{x-x^2}[/itex])
Vd = ∏ [itex]\sqrt{x-x^2}[/itex] dx
Vt = ∫∏[itex]\sqrt{x-x^2}[/itex] dx (upper limit 1, lower limit 0).
To find the volume o this sphere, I have to solve this integration problem:
∫∏[itex]\sqrt{x-x^2}[/itex] dx (upper limit 1, lower limit 0)
Here's where I get a problem. I attempt to use the substitution method, and I get this equation:
u = x-##x^2##
∫[itex]\sqrt{u}[/itex] dx
∫##u^(1/2)## dx
This equals [itex]\frac{u^(3/2)}{\frac{3}{2}}[/itex]
And I substitute u for x and end up with the final equation:
[itex]\frac{(x-x^2)^(3/2)}{\frac{3}{2}}[/itex]
When I set in limits for this equation and solve the problem, I get 0 - this is obviously wrong.
I'm fairly sure that I've integrated the problem wrong, but have I even attempted to integrate the correct expression?
Any help as always highly appreciated.
Edit: I have trouble with the formatting when I want to raise expressions to a higher power when I use latex.