a) Find the area enclosed between y=1/sqrt(1-x^2), the x axis and the lines x=0 and x=1/2.
integrating y I got [sin^-1(x)] between limits 1/2 and 0 which equals
b) This area is rotated about the y-axis. Find the exact volume of the soild formed.
This is what I am thinking...........use V= pi integral x^2 dy with limits pi/6 and 0. I rearranged y from part a) to make x^2 the subject and got x^2=1-y^-2.
Anywho I integrated and got V=5pi/4 units^3
The answer is pi/6(7sqrt(3)-12)units^3
I am way off, can someone please show me where I'm going wrong.
Basically, it comes from the definition of the volume of a cylinder: .
In our case, (or ) is the height of each shell (its usually the function you're considering rotating about a certain axis or line. In our case, its ). (or ) is the radius "distance" from the axis of rotation.
Since we want to revolve around the y axis, we measure the radius of each shell from 0 to 1/2 (in otherwords, for .
Now when we put this all together, we have , which is the integral I gave you.
I hope this clarifies how I came up with that integral.
See this pdf for more of an explanation on cylindrical shells.