Finding volume using cross sections

[miglo]

Homework Statement

find the volume of the solid whose base is the region bounded by y^2=9x and x=1 in the xy-plane. each cross section perpendicular to the x-axis is an isosceles right triangle with its hypotenuse in the xy-plane.

Homework Equations

V=\int_{a}^{b}A(x)dx

The Attempt at a Solution

well first i solved for y and got y=3sqrt(x) and y=-3sqrt(x), then i figured out the length of the hypotenuse to be 3sqrt(x)-(-3sqrt(x))=6sqrt(x)
but the formula for the area of an isosceles right triangle is 1/2a^2 where a is one of the triangles legs. how do i figure out the length of one of the legs?

 

[Mark44]

Homework Statement

find the volume of the solid whose base is the region bounded by y^2=9x and x=1 in the xy-plane. each cross section perpendicular to the x-axis is an isosceles right triangle with its hypotenuse in the xy-plane.

Homework Equations

V=\int_{a}^{b}A(x)dx

The Attempt at a Solution

well first i solved for y and got y=3sqrt(x) and y=-3sqrt(x), then i figured out the length of the hypotenuse to be 3sqrt(x)-(-3sqrt(x))=6sqrt(x)
but the formula for the area of an isosceles right triangle is 1/2a^2 where a is one of the triangles legs. how do i figure out the length of one of the legs?

I hope you have drawn sketches of everything, especially the triangles. The base of each triangle is 2y = 6√x. Each triangle is made up of two isosceles right triangles. The base of each is y = 3√x. Is that enough to get you started?

 

[miglo]

yeah got it, thanks!

 

[miglo]

just a quick question to check if i got the right answer, the area will then be 1/2(3sqrt(2x))^2=1/2*9*2x=9x
and integrating i get a volume of 9/2 since my limits will be from 0 to 1?