posted by Drake .
R is the region in the plane bounded below by the curve y=x^2 and above by the line y=1.
(a) Set up and evaluate an integral that gives the area of R.
(b) A solid has base R and the cross-sections of the solid perpendicular to the y-axis are squares. Find the volume of the solid.
(c) A solid has base R and the cross-sections of the solid perpendicular to the y-axis are equilateral triangles. Find the volume of the solid.
(a) The y = x^2 curve crosses y = 1 at x = 1. For the area in question, compute the integral of x^2 from x = 0 to 1, and subtract it from 1.
(b)Integrate along the y axis, with differential slab volume x^2 dy = y dy, from y=0 to y=1
(c) Proceed similary to (b), integrating along y from 0 to 1, but with equilateral triangle slabs. Slab volume must be expressed in terms of x
Why would it only be the integral of x^2 and not the integral of 1-x^2 from x=-1 to x=1? So wouldn't that equal an area of 4/3? And secondly, how would you sketch these solids? when you say integrate along the y-axis do you mean to say that the curves have to be shifted as well? And why are all the integrals from x=0 to x=1?
Find the volume, V , of the solid obtained
by rotating the region bounded by the graphs
x = y2, x = squareroot(y)
about the line x = −1.