Cross sections of a region

**amiv4** · October 2nd 2008, 06:45 PM

Find the volume of the described solid S

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

Some help please

**amiv4** · October 2nd 2008, 08:18 PM

i think all i need is the formula for the isosceles right triangle. So can anyone tell me what taht is. and would the integral go from -6 to 6

**icemanfan** · October 2nd 2008, 08:25 PM

Given $9x^2 + 4y^2 = 36$ , you have

$4y^2 = 36 - 9x^2$

$y^2 = \frac{36 - 9x^2}{4} = 9 - \frac{9x^2}{4}$

Now, the length of each triangle base is 2y.

So the area of each triangle is $\frac{1}{2}y\sqrt{2}\cdot y\sqrt{2} = y^2$ , which is $9 - \frac{9x^2}{4}$ .

Can you do the rest?

**amiv4** · October 2nd 2008, 08:26 PM

what would the integral be evaluated from

**icemanfan** · October 2nd 2008, 08:28 PM

Originally Posted by amiv4

what would the integral be evaluated from

What are the extreme left and right points on the ellipse?

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