# Volume of revolution between y=sinx and y=cosx

• May 28th 2009, 08:50 PM
drew.walker
Volume of revolution between y=sinx and y=cosx
How would I go about solving the following problem?

Find the volume of revolution defined by rotating, around the x-axis, the area between the graphs of y = sinx and y = cosx on the interval where 0 is less than or equal to x, which is less than or equal to 1.

I'm aware of the formula for calculating solids of revolution, and could solve the problem if it was just the area between y = sinx and the x-axis. I'm also aware of the "washer" method. However, the problem I'm having is that y = sinx and y = cosx cross over each other between 0 and 1, so I don't know how I'm supposed to deduce the formulas for the radius of the inner and outer circle of each washer. Should I be treating them as two different solids of revolution on an interval of 0 to the crossover point and the crossover point to 1? I think this would work, but I doubt it's the correct/best method. Any help would be greatly appreciated.
• May 28th 2009, 09:42 PM
TheEmptySet
Quote:

Originally Posted by drew.walker
How would I go about solving the following problem?

Find the volume of revolution defined by rotating, around the x-axis, the area between the graphs of y = sinx and y = cosx on the interval where 0 is less than or equal to x, which is less than or equal to 1.

I'm aware of the formula for calculating solids of revolution, and could solve the problem if it was just the area between y = sinx and the x-axis. I'm also aware of the "washer" method. However, the problem I'm having is that y = sinx and y = cosx cross over each other between 0 and 1, so I don't know how I'm supposed to deduce the formulas for the radius of the inner and outer circle of each washer. Should I be treating them as two different solids of revolution on an interval of 0 to the crossover point and the crossover point to 1? I think this would work, but I doubt it's the correct/best method. Any help would be greatly appreciated.

See the attached graph

Attachment 11660

So from zero to $\displaystyle \frac{\pi}{4}$ the cosine is the upper function and then sine is from $\displaystyle \frac{\pi}{4}$ to 1.

$\displaystyle \pi\int_{0}^{\frac{\pi}{4}}\cos^2(x)-\sin^2(x)dx+\pi\int_{\frac{\pi}{4}}^{1} \sin^2(x)-\cos^2(x)dx$

$\displaystyle \pi\int_{0}^{\frac{\pi}{4}}(\cos(x)-\sin(x))(\cos(x)+\sin(x))dx+\pi\int_{\frac{\pi}{4} }^{1} (\sin(x)-\cos(x))(\sin(x)+\cos(x))dx$

$\displaystyle \pi \frac{(\cos(x)+\sin(x))^2}{2} \bigg|_{0}^{\pi/4}+\pi \frac{(\sin(x)-\cos(x))^2}{2} \bigg|_{\pi/4}^{1}$

$\displaystyle \frac{\pi}{2}((\sqrt{2})^2-1^2+(\sin(1)-\cos(1))^2-(0))$

$\displaystyle \frac{\pi}{2}(2-1+\sin^2(1)-2\sin(1)\cos(1)+\cos^2(1))=\frac{\pi}{2}(2-2\sin(1)\cos(1))=\pi(1-\sin(1)\cos(1))$
• May 29th 2009, 02:49 AM
drew.walker
That's great! My only question about your working would be: how did you know that the graphs were equal at pi/4? Was this something you calculated? If so, how?
• May 29th 2009, 03:40 AM
mr fantastic
Quote:

Originally Posted by drew.walker
That's great! My only question about your working would be: how did you know that the graphs were equal at pi/4? Was this something you calculated? If so, how?

Solve $\displaystyle \sin x = \cos x \Rightarrow \tan x = 1$.