# Math Help - Cross sections of a region

1. ## Find the volume of the described solid

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.

2. 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

3. 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?

4. what would the integral be evaluated from

5. Originally Posted by amiv4
what would the integral be evaluated from
What are the extreme left and right points on the ellipse?