# Volumes of Regions with Known Cross Sections

• November 2nd 2009, 11:27 PM
RockHard
Volumes of Regions with Known Cross Sections
Find the volume of the solid bounded by the curves y = x^2 and y = 8 - x^2 who cross sections are perpindicular to the x - axis with one side on the xy plane using semi circle cross sections

Ok so i have to find total volume which is equal to the Rheiman sum of the the volume of the semi circle which is

$\frac{1}{2}\pi*r^2*\delta thickness$

and where delta thickness is the change in x?
and the radius of the semi circle is 1/2 the height between y = 8 - x^2 and y = x^2
which gives

$\frac{8-2x^2}{2}$

$4-x^2$

then you will have the intergral from x = -2 to x = 2 because this is where they intersect

$\frac{\pi}{2}\int{4-x^2}^2$

For some reason I dont think im accurately doing this
• November 3rd 2009, 03:15 AM
HallsofIvy
Quote:

Originally Posted by RockHard
Find the volume of the solid bounded by the curves y = x^2 and y = 8 - x^2 who cross sections are perpindicular to the x - axis with one side on the xy plane using semi circle cross sections

Ok so i have to find total volume which is equal to the Rheiman sum of the the volume of the semi circle which is

$\frac{1}{2}\pi*r^2*\delta thickness$

and where delta thickness is the change in x?

Yes, because your cross sections are "perpendicular to the x-axis", the thickness, perpendicular to the area, is measured along the x-axis.

Quote:

and the radius of the semi circle is 1/2 the height between y = 8 - x^2 and y = x^2
which gives

$\frac{8-2x^2}{2}$

$4-x^2$

then you will have the intergral from x = -2 to x = 2 because this is where they intersect

$\frac{\pi}{2}\int{4-x^2}^2$

For some reason I dont think im accurately doing this
Two errors:
You left out the "thickness" which becomes "dx" in the limit. NEVER write an integral without the "dx"!
Also, it is the entire polynomial that is squared. $\frac{\pi}{2}\int \left(4- x^2\right)^2 dx$.

Now, what are the limits of integration?
• November 3rd 2009, 01:35 PM
RockHard
The limits of integration are lower limit = -2 and upper limit = 2 because thats where the two graphs intersect?
• November 4th 2009, 10:05 AM
RockHard
Bump
• November 5th 2009, 07:25 AM
RockHard
...again..bump