Volume of Solid w/ Elliptical Base & Right Triangles

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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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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:
 
Find volume via method of cross-sections.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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