- #1
zeion
- 466
- 1
Volume by cross-section: ellipse and equilateral triangle cross sections??
The base of a solid is the region bounded by the ellipse 4x^2+9y^2=36. Find the volume of the solid given that cross sections perpendicular to the x-axis are:
a) equilateral triangles
b) squares
So I'm not really sure how ellipses work.. how can I sketch this ellipse?
Beyond that.. I try to calculate the area of the triangle and then integrate in terms of y so the base is changing according to the ellipse curve.
I write the ellipse as:
y = +/-sqrt((-4/9)x^2 + 4)
So the base of the triangle is 2(sqrt((-4/9)x^2 + 4))
And has that as the length on all side since it is equilateral.
Then I try to find the height using Pythagoras and get
h = +/-sqrt((-4/3)x^2 + 12)
Then now I have the area of the triangle as (1/2)bh, which is =
A = (1/2)(2(sqrt((-4/9)x^2 + 4)))(sqrt((-4/3)x^2 + 12))
Then I can integrate in terms of x.. does that look correct so far?
Homework Statement
The base of a solid is the region bounded by the ellipse 4x^2+9y^2=36. Find the volume of the solid given that cross sections perpendicular to the x-axis are:
a) equilateral triangles
b) squares
Homework Equations
The Attempt at a Solution
So I'm not really sure how ellipses work.. how can I sketch this ellipse?
Beyond that.. I try to calculate the area of the triangle and then integrate in terms of y so the base is changing according to the ellipse curve.
I write the ellipse as:
y = +/-sqrt((-4/9)x^2 + 4)
So the base of the triangle is 2(sqrt((-4/9)x^2 + 4))
And has that as the length on all side since it is equilateral.
Then I try to find the height using Pythagoras and get
h = +/-sqrt((-4/3)x^2 + 12)
Then now I have the area of the triangle as (1/2)bh, which is =
A = (1/2)(2(sqrt((-4/9)x^2 + 4)))(sqrt((-4/3)x^2 + 12))
Then I can integrate in terms of x.. does that look correct so far?