Recent content by Alixx69

For the triangle cross sections V = A*h h = dx A=.*l*w l = w = sqrt(1-x^2) is that right? would that make the integral V=int from -1 to 1(.5(2*sqrt(1-x^2)))? that gives me an answer of .5pi but I am so sure the answer is pi because vv how can it not be half a cylinder, if you cut a cylinder...

i meant the height being being height and the base being the area of the circle, but what you said got me thinking... so the height being the value y of the solved equation x^2+y^2=r^2(r is the radius) which is y=sqrt(r^2-x^2) and the base being the area of the circle which is pi*r^2 then it...

i used r(radius of the circle) because i was solving the formula for the area of a circle, x^2+y^2=r^2, for y, which is sqrt(r^2-x^2) is that correct? in my book it shows the answer to be found by V = int from 0 to 2(2Xsqrt(2*r*x-x^2))dx but i have no idea how they got there, how did they? and...

yeah i realized that and made a new thread, but so what your saying is that the base isn't x, using x^2+y^2=r^2 and the fact that using x^2+y^2=1 is the area equation for the base that the base is sqrt(2r^2+x^2)? is that the height as well?

[[URGENT]]Vof solid with circular base&right isoceloes perpendicular cross sections Homework Statement 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...

i have a problem that is almost identical to bluebear19's so i thought it would be appropriate to post it on this thread, since, Dick, you seem to know it very well 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...