Volume By Slicing

• Aug 3rd 2010, 10:32 PM
dreamx20
Volume By Slicing
The question is read as follows: Solid lies between planes perpendicular to the x-axis at x=1 and x=-1 . The cross sections are perpendicular to the x-axis are circular disk.s whose diamters run from the parabola y=x^2 to the parabola y=2-x^2 FInd the volume of the solid.

I tried doing this by the given integral - v= int (a to b) A(x) dx =int (a to b) pi R(x)^2 dx
I ve determined that the radius should equal to 2 . I dont know where to go from here to determine the limits of integration (a and b values) and such. Any help is appreciated :D
• Aug 4th 2010, 01:51 AM
earboth
Quote:

Originally Posted by dreamx20
The question is read as follows: Solid lies between planes perpendicular to the x-axis at x=1 and x=-1 . The cross sections are perpendicular to the x-axis are circular disk.s whose diamters run from the parabola y=x^2 to the parabola y=2-x^2 FInd the volume of the solid.

I tried doing this by the given integral - v= int (a to b) A(x) dx =int (a to b) pi R(x)^2 dx
I ve determined that the radius should equal to 2 . I dont know where to go from here to determine the limits of integration (a and b values) and such. Any help is appreciated :D

1. Draw a sketch of the situation.

2. The midpoints of the disks are placed on the line y = 1. Thus the radius of the disks can be calculated by

$r = 1-x^2$

3. The limits of integration are obviously given by the x-coordinates of the points of intersection between the two graphs.

4. I've got a volume of $V=\frac{16}{15} \pi \ vol.units$
• Aug 4th 2010, 02:39 AM
HallsofIvy
Since, as earboth says, the radius of each disk is $1- x^2$, the area is $\pi(1- x^2)^2= \pi(1- 2x^2+ x^4)$. The thickness of each disk is along the x-axis and so is "dx". The volume of a disk is "area times thickness", $\pi(1- 2x^2+ x^4)dx$.

"Add" the thicknesses of all the disks. In the limit, that becomes the integral $\pi\int_{-1}^1 1- 2x^2+ x^4 dx$.