Help with Volume circular disk.

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SUMMARY

The volume of the solid S, which has a circular disk base with a radius of 3r and square cross-sections perpendicular to the base, is calculated using integration. The correct approach involves determining the length of the chord at a distance x from the center of the circle, which forms the side length of the square cross-section. The area A(x) is derived from the chord length, and the volume V is computed by integrating A(x) from 0 to 3r and multiplying the result by 2. The final volume is expressed as V = 18πr³.

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quickclick330
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Consider the solid S described below.
The base of S is a circular disk with radius 3r. Parallel cross-sections perpendicular to the base are squares.
Find the volume V of this solid.


I tried this...


A(x) = pi*r^2 = pi*(3r)^2 = pi*9r^2

V(x) = integral from 0 to 3r(pi*9r^2 dx) = pi*27r^3


Am I approaching this problem the wrong way? Thanks for the help!
 
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quickclick330 said:
Consider the solid S described below.
The base of S is a circular disk with radius 3r. Parallel cross-sections perpendicular to the base are squares.
Find the volume V of this solid.

The square cross-sections are built upon the circular base, so each cross-section has a side length given by the length of a chord of the circle; these "slices" are standing upright on the circle. (Making a picture of this will be helpful to you.) The solid is symmetric about a diameter of the circle, so you can integrate half the solid and multiply your result by two.

So what you need is to find the length of a chord of the circle which has its midpoint at a distance x from the circle's center. That length gives you the area A(x) of each cross-section. You would then integrate A(x) from x = 0 to x = 3r and multiply the resulting volume by 2.
 

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