# Thread:

1. ## Solid of revolution

The area under the curve y= sin x, between x =0 and x = π
Show that the volume of the solid formed by rotating this area through 360 about the x-axis is exactly ½(π^2)

This is what I have after doing some of the work:
V = π (-cos^2x) limits π and 0
I am not sure how to take it any further to obtain the answer. Please help me out guys...

2. Originally Posted by mike789
The area under the curve y= sin x, between x =0 and x = π
Show that the volume of the solid formed by rotating this area through 360 about the x-axis is exactly ½(π^2)

This is what I have after doing some of the work:
V = π (-cos^2x) limits π and 0
I am not sure how to take it any further to obtain the answer. Please help me out guys...
your antiderivative is incorrect ... use the power reduction identity that follows.

$\displaystyle V = \pi \int_0^\pi \sin^2{x} \, dx$

$\displaystyle V = \int_0^\pi \frac{1-\cos(2x)}{2} \, dx$

now try it again.

3. Originally Posted by skeeter
your antiderivative is incorrect ... use the power reduction identity that follows.

$\displaystyle V = \pi \int_0^\pi \sin^2{x} \, dx$

$\displaystyle V = \int_0^\pi \frac{1-\cos(2x)}{2} \, dx$

now try it again.
Do i need to integrate and substitute the limits?

4. Originally Posted by mike789
Do i need to integrate and substitute the limits?
if that's what you mean by evaluating the definite integral using the Fundamental Theorem of Calculus, then yes.

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