# Finding the volume of the solid generated by revolving the area

• stardust006
In summary, the volume of the solid generated by revolving the area bounded by the given curve about the indicated axis is 6 cubic units.
stardust006
find the volume of the solid generated by revolving the area bounded by the given curve about the indicated axis:y = 2x-x^2 and y = x; about the y-axis
The attempt at a solution:

so i assigned values for x and y, and the curve is a parabola that opens downward with a vertex of (1,1), i used cylindrical shell method of integral calculus and got an equation of,
2∏∫from 0 to 1 of (x) (x-2x+x^2) dx.

Is this right?

You have the correct idea, subtracting one function from the other. If you evaluate your integral what answer do you get, does it make sense?

I'm hinting at the fact that you subtracted in the wrong order. In problems like these, you want to subtract the "smaller" function from the "bigger." In this case, $2x-x^{2} \geq x$ for $y \in [0,1]$.

What you are integrating over is the area between these two functions. In order to get that area, you subtract the lower function from the bigger to trim off the extra area below the area of integration. Does that make sense?

It seems like you need to flip the sign, otherwise it's negative. Other than that it looks right.

Instead of just flipping the sign, set up your typical area element so that its area is a positive number. Since y = 2x - x2 is above the line y = x, the height of the typical area element is 2x - x2 - x, not x - 2x - x2. If you make that change, you'll get a positive number for the volume of the rotated region.

stardust006 - do not start multiple posts for the same problem. I merged one reply from the other thread into this thread.

@mark44, sorry..

thanks guys, so is the answer ∏/6 cubic units?

Ok, thanks again! :)

## 1. What is the formula for finding the volume of a solid generated by revolving the area?

The formula for finding the volume of a solid generated by revolving the area is V = ∫abπ[R(x)]2dx, where a and b are the limits of integration, and R(x) is the radius of the area being revolved.

## 2. How do you determine the limits of integration when finding the volume of a solid generated by revolving the area?

The limits of integration are determined by the boundaries of the area being revolved. These can be found by setting up an integral for the area and solving for the points where it intersects the x-axis.

## 3. Can the volume of a solid generated by revolving the area be negative?

No, the volume of a solid generated by revolving the area cannot be negative. It is a physical quantity and therefore must be positive or zero.

## 4. What is the difference between revolving the area around the x-axis and the y-axis?

When revolving around the x-axis, the radius of the area is a function of x, while when revolving around the y-axis, the radius is a function of y. This will result in different integrals and limits of integration.

## 5. How do you know if you should use the disc method or the shell method when finding the volume of a solid generated by revolving the area?

The disc method is used when the area is revolved around an axis perpendicular to the plane of the area, while the shell method is used when the area is revolved around an axis parallel to the plane of the area. Therefore, the method to use depends on the orientation of the axis of revolution.

Replies
1
Views
1K
Replies
8
Views
2K
Replies
1
Views
1K
Replies
4
Views
1K
Replies
3
Views
1K
Replies
12
Views
4K
Replies
2
Views
1K
Replies
2
Views
730
Replies
9
Views
2K
Replies
8
Views
2K