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. find the volume of the described solid s?

To obtain the equation for an elliptical region, we divide the boundary curve equation by 36, which gives us:

[9x^2 + 4y^2 = 36] / 36

(x^2)/4 + (y^2)/9 = 1

Since the given cross-sections are isosceles right triangles, the hypotenuse is found on the base. Using trigonometric functions, the hypotenuse is found to be 6 * sqrt(1 - (x^2)/4). The cross-sectional area is then found to be  9 (1 - (x^2)/4).

With the cross-section, we integrate it with limits of -2 to 2 in terms of x to find the volume. This is shown below:

 ∫ 9 (1 - (x^2)/4) dx (-2,2) = 24

Therefore, the volume is 24 units^3.

4+2=-2. 1+2=3. y=-2/3x+b. 2=4/3+b. 2/3=b. the answer is y= -2/3x+2/3