Find the volume of the solid (strange one )

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Homework Help Overview

The problem involves finding the volume of a solid whose base is defined by the circle x² + y² = 1, with cross sections perpendicular to the x-axis being semi-circles. Participants are exploring how to set up the integral for calculating the volume based on these parameters.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between the radius of the semi-circles and the equation of the circle, questioning how to express the radius in terms of x. There is also consideration of the integral setup and the bounds for integration.

Discussion Status

The discussion is active, with participants offering different perspectives on visualizing the solid and setting up the integral. Some guidance has been provided regarding the relationship between the radius and the variable x, but no consensus on the final approach has been reached.

Contextual Notes

Participants are grappling with the implications of the problem's constraints, such as the fixed radius of the base and the variable radius of the semi-circular cross sections. There is mention of a solution manual, but its contents are not disclosed in the discussion.

Saladsamurai
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...whose base is enclosed by the circle [tex]x^2+y^2=1[/tex] and whose cross sections taken perpendicular to the x-axis are semi-circles.

So I understand that the volume of each "sample" semi-circle will be [tex]\frac{\pi r^2}{2}*\Delta x[/tex]

The radius is y. Now this is where I lose the solution manual...Do I need to take into account that [tex]y=^+_-\sqrt{1-x^2}[/tex] ? I am having trouble setting up this integral.

I am working on the bounds now...but^^^^ that part is effing me up.

Blahhhh,
Casey
 
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[tex]V = \pi \int y^2 dx[/tex]

Just let [itex]y^2 = r^2 - x^2[/itex] where r is a constant, and then its just a case of picking the limits of the integral to suit the problem. The integral itself shouldn't present any problems.
 
Well, the radius IS y..so I don't follow you.

Casey
 
I'm trying to picture this. It just seems like its half a sphere, or a dome shape. Maybe I'm visualizing it wrong. If I'm right, they could have just said a dome.

If I'm right and you want to do it without calculus, its just the formula for a sphere divided by 2: 2*pi*r^3/3. Assuming I'm picturing the right shape, you could at least use this to check your answer if you use integration. What does the solution manual say is the answer? Does it match 2*pi*r^3/3, which for r=1 would be 2*pi/3?

In your example, becareful because there are 2 r's you're dealing with, one is a fixed value which is the radius of the base which = 1 from the question. The r in your integral symbol is a variable representing the radius of a slice.

So substitute your r in the integral with the sqrt(1-x^2).
 
Saladsamurai said:
Well, the radius IS y..so I don't follow you.

Casey
Yes, and since the "thickness" is [itex]\Delta x[/itex] you will need to write y2 as a function of x. That's easy!
 
I got it. It's just twice the volume of the "half-dome' whose base is the half-circle y=(1-x^2)^.5 from 0 to 1...like tony said! Sometimes I just need to move on an look at the problem again later (whenI have gotten some sleep:/ )
 

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