Volume of Solid w/ Elliptical Base & Right Triangles

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In summary, the given solid has a base of an elliptical region with boundary curve 9x^2 + 4y^2 = 36. The cross-sections perpendicular to the x-axis are isosceles right triangles with the hypotenuse in the base. To solve, use the method of cross-sections and the fact that the base is 2y and the height can be calculated as it is a right-angled isoceles triangle.
  • #1
vigintitres
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Homework Statement


The base of S is an elliptical region with boundary curve 9x^2 + 4y^2 = 36. Cross-sections perpindicular to the x-axis are isosceles right triangles with hypotenuse in the base.

Find the volume of the described solid.


Homework Equations


V = {int} 1/2 b*h dy



The Attempt at a Solution


I found that I would have to use the symmetry to solve this. The only things I have are x^2 + y^2 = 1/2 and y = sqrt(.5 - x^2)

Now i know I have to integrate an isosceles triangles area which is 1/2 b*h but I'm not sure what the base or the height will be.
 
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  • #2
Hi vigintitres! :smile:
vigintitres said:
The base of S is an elliptical region with boundary curve 9x^2 + 4y^2 = 36. Cross-sections perpindicular to the x-axis are isosceles right triangles with hypotenuse in the base.

Find the volume of the described solid.

I found that I would have to use the symmetry to solve this. The only things I have are x^2 + y^2 = 1/2 and y = sqrt(.5 - x^2)

Where do you get x^2 + y^2 = 1/2 from? :confused:

9x^2 + 4y^2 = 36.
Now i know I have to integrate an isosceles triangles area which is 1/2 b*h but I'm not sure what the base or the height will be.

The base is 2y.

The height you can work out because it's a right-angled isoceles triangle. :wink:
 
  • #3
Find volume via method of cross-sections.
 

1. What is the formula for finding the volume of a solid with an elliptical base and right triangles?

The formula for finding the volume of a solid with an elliptical base and right triangles is V = (π * a * b * h)/3, where a and b are the lengths of the semi-major and semi-minor axes of the ellipse, and h is the height of the solid.

2. How is an elliptical base different from a circular base when finding volume?

An elliptical base is different from a circular base when finding volume because the formula for the volume of a cylinder (π * r^2 * h) only applies to a circular base. For an elliptical base, the formula must take into account the lengths of both the semi-major and semi-minor axes of the ellipse.

3. Can the volume of a solid with an elliptical base and right triangles be found using calculus?

Yes, the volume of a solid with an elliptical base and right triangles can be found using calculus. The solid can be divided into infinitesimally small cones, and the volume of each cone can be calculated using the formula for the volume of a cone (π * r^2 * h/3). The volumes of these cones can then be integrated to find the total volume of the solid.

4. Are there any real-life applications of finding the volume of a solid with an elliptical base and right triangles?

Yes, there are many real-life applications of finding the volume of a solid with an elliptical base and right triangles. For example, this formula can be used to calculate the volume of a water tank with an elliptical base, or the volume of a storage container with a triangular cross-section. It can also be used in engineering and construction for designing buildings or structures with unique shapes.

5. Can the volume of a solid with an elliptical base and right triangles be negative?

No, the volume of a solid with an elliptical base and right triangles cannot be negative. Volume is a measure of the amount of space occupied by an object, and it cannot be negative. If the result of the formula is negative, it is likely due to a miscalculation or incorrect input of the dimensions.

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