Finding the volume of the solid generated by revolving the area

[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?

 

[scurty]

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?

 

[sunjin09]

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

 

[Mark44]

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.

 

[Mark44]

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

 

[stardust006]

@mark44, sorry..

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

 

[scurty]

That's the answer I got!

 

[stardust006]

Ok, thanks again! :)