• Support PF! Buy your school textbooks, materials and every day products Here!

Extreme confusion with volume Calculus

  • Thread starter flyingpig
  • Start date
  • #1
2,571
1

Homework Statement




Find the volume of a solid with a circular base defined by [tex]x^2 + y^2 = 1[/tex] with parallel cross sections perpendicular to the base are equilateral triangles.


The Solution

[tex]A(x) = \frac{1}{2} \cdot 2\sqrt{1-x^2} * \sqrt{3}\sqrt{1-x^2} = \sqrt{3}(1-x^2)[/tex]

[tex]\int_{-1}^{1} \sqrt{3}(1-x^2) dx = \frac{4\sqrt{3}}{3}[/tex]

Question

Why is the height only [tex] \sqrt{3}\sqrt{1-x^2}[/tex] and not [tex]2 \sqrt{3}\sqrt{1-x^2}[/tex]? I thought [tex]\sqrt{1-x^2}[/tex] is only the half the height of the circle
 

Answers and Replies

  • #2
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,545
756
y = sqrt(1-x2) is half the base of the triangle so

b = 2y, h = y sqrt(3)

A = (1/2)bh = (1/2)(2y)(ysqrt(3) = y2sqrt(3) = (1-x2)sqrt(3)
 
  • #3
2,571
1
No I am asking about the height not the base.
 
  • #4
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,545
756
No I am asking about the height not the base.
Did you actually read what I wrote? Guess what h stands for.
 
  • #5
2,571
1
h stands for height, but that's not the nature of the question. y = sqrt(1 + x^2) which is only half the height
 
  • #6
I like Serena
Homework Helper
6,577
176
Question

Why is the height only [tex] \sqrt{3}\sqrt{1-x^2}[/tex] and not [tex]2 \sqrt{3}\sqrt{1-x^2}[/tex]? I thought [tex]\sqrt{1-x^2}[/tex] is only the half the height of the circle
The height of a unit equilateral triangle is [tex]\frac 1 2 \sqrt{3}[/tex]
The side of the triangle would be, as you state, [tex]2 \sqrt{1-x^2}[/tex].
Note that in the product of the two the 1/2 and the 2 cancel against each other.
 

Related Threads on Extreme confusion with volume Calculus

  • Last Post
Replies
7
Views
1K
Replies
10
Views
1K
Replies
0
Views
3K
Replies
1
Views
2K
Replies
8
Views
770
Replies
7
Views
5K
Replies
1
Views
521
Top