Volume of Sold with a Known Crosssection

[mmg0789]

the base is bounded by x^2 + y^2 = 4
the solid is formed by isosceles right triangles perpendicular to x axis. the hypotenuse of each triangle is sitting on the base of the figure

i just need to know what area formula i would start out with, is it just bh/2 ?

 

[HallsofIvy]

An "isosceles right triangle perpendicular to the x axis" has hypotenuse y. Since it is isosceles, a²+ a²= 2a²= y². That is, each leg has length $a= y\sqrt{2}$ and so its area is (1/2)bh= (1/2)(a*a)= (1/2)(2y)= y.

 

[mmg0789]

hmm i found that a= sqrt(1/2) * y from 2a^2=y^2, the area would then be
(1/4)y^2. Right?

 

[MalayInd]

hmm i found that a= sqrt(1/2) * y from 2a^2=y^2, the area would then be
(1/4)y^2. Right?

yes, you are right.