Volume of solid of revolution - y axis.

[Mutaja]

Homework Statement

Find the volume of the solid of revolution when we rotate the area limited by the x-axis and the function f(x) = 1 - cosx where x e [0, 2∏] once around the y-axis?

The Attempt at a Solution

In my notes I have the following equation:

V = ∫ 2∏x f(x) dx

If I put in my limits (upper limit 2∏, lower limit 0) and my function I get the following:

V = 2∏ ∫x(1-cos(x)) dx

V = 2∏ ∫x - xcos(x) dx

V = 2∏[\frac{x^2}{2} - (xsin(x)+cos(x))]

V = 2∏ [\frac{x^2}{2} - (∏sin(∏) + cos(∏)] - 2∏ [\frac{0^2}{2} - (0sin(0) + cos(0)]

Since ∏ sin(∏) = 0, cos(∏) = -1 , 0sin(0) = 0 and cos(0) = 1 I get the following:

V = 2∏ (\frac{∏^2}{2}) - 2∏ + 1

V = $2∏^3$ - 4∏ + 2

Is this correct? Am I using the correct formulas/equations?

Please let me know if there is something I need to explain better. Any help and guiding is massively appreciated. Thanks.

 

[PeroK]

V = 2∏ ∫x - xcos(x) dx

V = 2∏[\frac{x^2}{2} - xsin(x)+cos(x)]

Check your integration by parts: should be -cos(x).

 

[Mutaja]

Check your integration by parts: should be -cos(x).

Sloppy mistake by me, there should of course be parenthesis around that equation.

V = 2∏[\frac{x^2}{2} - (xsin(x)+cos(x))]

That slipped past me when I was writing off of my notes - therefore unless I again have overlooked something (which I've double checked I haven't), that mistake was a one off - I've solved the rest of the problem as if there were parenthesis around (xsin(x)+cos(x)). Therefore I get - (-1) -> +1 in my answer which I later multiply by 2.

Does it look ok besides that error?

Thanks for your input!