2.  Find the volume of y= X^2, y=4 revolved around the x-axis  Cross sections are circular washers  Thickness of the washer is xsub2-xsub1  Step 1) Find the thickness. (Volume of entire)^2 – (Volume of the “hole”)^2. In this case it is (4)^2-(X^2)^2

3.  Step 2) Find the bounds. Because y=4 and y=x^2 intersect at x=-2 and x=2, those are your bounds.  Step 3) Plug all the information into the integral  It should like this :  from x=-2 to x=2 of (4)^2- (X^2)^2 all multiplied by pie  The answer should be : 256pie/5

4.  Find the volume of y=X^2 between x=0 and x=1 revolved around the x-axis  Step 1) For this problem, you must do “top- bottom” to find the “thickness”, so it is ((x^2)-0)  Step 2) Next, you must find the bounds. As given in the problem, the bounds are from x=0 to x=1

5.  Step 3) Set up the integral from x=0 to x=1 and square the thickness of the disk, which is ((x^2)-0)  In every disk problem, you will need to square the “thickness”  Step 4) multiply everything by pie.  It should look like this:  from x=0 to x=1, ((x^2)-0)^2 dx all multiplied by pie  The answer should be: pie/5

6.  Find the volume of the solid generated when the curve y=x^2 between x=1, x=2 is rotated around the y-axis  Step 1) Sketch a line segment parallel to axis of revolution. This is the height of a cylinder.

7.  Step 2) Connect this segment perpendicular to the axis of revolution. This is the radius of a cylinder.  Step 3)Find the limits of integration. If rotating around x-axis, then it will be dy. If rotating around y-axis, then it will be dx.  Step 4) Integrate using Surface Area= (2)(r)(pie)

8.  So, it would be: 2pie from 1 to 2 of (radius)(height) dx radius=(x) & height=(x^2-0) The answer should be :15pie/2