12. The base of a solid is a circular disk with radius of 6. Find the volume of the solid...

Question

12. The base of a solid is a circular disk with radius of 6. Find the volume of the solid if parallel cross sections perpendicular to the x-axis are squares. Set-up but do not evaluate the integral.

 

 

ANSWER: 

2 ∫ (2 √(36 - x²) )² dx

a = 0

b = 6

 

Here's what I DO understand:

*   Where √(36 - x²) comes from

*   Why we square the entire function that we're evaluating

*   Why it's dx, not dy

 

Here's what I DON'T understand:

(1)   Where the 2 outside the integral comes from

(2)   Where the 2 inside the integral comes from

(3)   Why he choose 0 and 6 as a and respectively. Would -3 and 3 work?

 

Thanks!

 

Answer 1

(1) & (3) are related to the fact that there's symmetry; insstead, if the limits a & b were -6 & 6, the outside factor 2 would not be needed.  [No, -3 to 3 won't work--see below]

 

Visualize by thinking of the base circle having equation x² + y² = 36 so you can solve for y².

The vertical cross section is a square, taken parallel to the y-axis and perpendicular to the x-axis, so one side of the square has 2y as its side length.  Therefore the cross-sectional area is (2y)² and this suggests a thin volume when multiplied by differential (dx).  Then you integrate from a = 0 to b = 6, to get the volume, which is shaped like a round loaf of French bread, I believe.  It's dx because you visualize these many slices as cutting through the x-axis (with a bread knife).

 

I would appreciate a feedback from you, commenting on whether you like this explanation.