Thread: 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...

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.





now try it again.

Originally Posted by skeeter

your antiderivative is incorrect ... use the power reduction identity that follows.





now try it again.

Do i need to integrate and substitute the limits?

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.