[[URGENT]]Vof solid with circular base&right isoceloes perpendicular cross sections 1. The problem statement, all variables and given/known data so there is a circle described by the equation x^2+y^2=1 with right isosceles triangles perpendicular to it. each of the triangles is touching the circle with one of its LEGS, with the 90 degree angle and one of the smaller angles touching the perimeter of the circle. now use an integral to find the volume of the solid produced by the adding all of the areas of the infinitely thin triangles from y = -1 to 1. (Just in case my interpretation wasn't as clear to you as it is to me, the actual question is: "The base of the solid is the disk x^2+y^2 <= 1. The cross-sections by planes perpendicular to the y axis between y= -1 and y=1 are isosceles right triangles with one LEG on the disk") 2. Relevant equations V=int from a to b(A(x)dx) 3. The attempt at a solution so i got, A(x)=int from -1 to 1(height of the triangle*base of the triangle) base = x and this is where i get tripped up, if the base, being one of the LEGS of the triangle is x, shouldnt the height(the other LEG of the isosceles triangle) be x too? that would make the answer: V(x)=int from -1 to 1((.5x^2)dx)... but that would mean that , using .5[(x^3)/3] as the antiderivitive and evaluating from -1 to 1, the answer is 1/3 but if you think about it in a different way , its a half of a can that has a r of 1 and a h of 2 and using (pi*r^2*h)/2 you would get 3.14 which is pi where am i going wrong with this?