# Thread: Isosceles right triangles perp to base...Help.

1. ## Isosceles right triangles perp to base...Help.

Base of the solid is the area located by the curve. It says cross-sections perpendicular to the x-axis are isosceles right triangles with hypotenuse in the base. I found the upper limit to be 3 and the lower limit to be -3. Can anyone show me how to find the base and the height? I used the 45/45/90 rule and came up with ((144-16x^2)/9)^(1/2)*((2)^(1/2)/2) for the hypotenuse. I know this is wrong, but now really sure how to actually find the base or height.

a=-3
b=3

2. Originally Posted by Bracketology
Base of the solid is the area located by the curve. It says cross-sections perpendicular to the x-axis are isosceles right triangles with hypotenuse in the base. I found the upper limit to be 3 and the lower limit to be -3. Can anyone show me how to find the base and the height? I used the 45/45/90 rule and came up with ((144-16x^2)/9)^(1/2)*((2)^(1/2)/2) for the hypotenuse. I know this is wrong, but now really sure how to actually find the base or height.

a=-3
b=3
hypotenuse = $2y$

area of a right isosceles triangle is $\frac{(hyp)^2}{4} = \frac{(2y)^2}{4} = y^2$

$\displaystyle V = 2 \int_0^3 y^2 \, dx
$

$\displaystyle V = 2 \int_0^3 \frac{144-16x^2}{9} \, dx$

$\displaystyle V = \frac{2}{9} \int_0^3 144-16x^2 \, dx$

3. Where did the 2y come from? How do you know the hypot. is 2y?

4. Originally Posted by Bracketology
Where did the 2y come from? How do you know the hypot. is 2y?
the vertical distance from the x-axis to the elliptical curve is y ... since the ellipse is symmetrical to both axes, the total vertical distance from the lower part of the ellipse to the upper part is 2y.