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

• Nov 7th 2010, 06:37 PM
Bracketology
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.

Attachment 19631

a=-3
b=3
• Nov 7th 2010, 06:48 PM
skeeter
Quote:

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.

Attachment 19631

a=-3
b=3

hypotenuse = $\displaystyle 2y$

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

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

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

$\displaystyle \displaystyle V = \frac{2}{9} \int_0^3 144-16x^2 \, dx$
• Nov 7th 2010, 06:49 PM
Bracketology
Where did the 2y come from? How do you know the hypot. is 2y?
• Nov 7th 2010, 06:56 PM
skeeter
Quote:

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.