# Solid of revolution

• Nov 1st 2010, 02:42 PM
mike789
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
• Nov 1st 2010, 03:08 PM
skeeter
Quote:

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

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.
• Nov 1st 2010, 04:16 PM
mike789
Quote:

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?
• Nov 1st 2010, 04:19 PM
skeeter
Quote:

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.