Find the volume of the solid given the cross sections perpendicular to the x-axis. Caluclus?

To do it you will need to work out the positions of the three corners of this triangle which is supposed to be the the base of each solid. They are the three crossing points of the lines x+2y=4, x-2y=4 and x=0, the last of these being the equation of the line congruent with the y-axis. Each corner lies on two of these lines. Their co-ordinates are therefore obtained by considering pairs of these equations simultaneously. The first two two taken together imply that x+2y+x-2y=4+4, so 2x=8 and so x=4 and therefore y=0. Taking the last two together, we get -2y=4, so y=-2 and we already have x=0. Taking first and last together, we get 2y=4, so y=2 and again, we already have x=0. The the co-ordinates of the three vertices are (4, 0), (0, -2) and (0, 2).

There is no need for calculus in this question if you can use the formulae for the volume of slant pyramids, which is base area multiplied by a third of the height.

In part a, the end of the solid forms a square in the y-z plane (formula x=0) and has square cross-sections vanishing to a point at (4,0) on the x-axis. Using this as the base (instead of the triangle given), we have a square-based slant pyramid of height 4 units. The edge of the end square along the y-axis is 4 units long so its area must be 16 square units. The volume of the solid is therefore 16*4/3, which is 64/3 cubic units.

For part b, it is better to use the triangle in the x-y plane as the base. We then have a triangular based slant pyramid with its apex on the z-axis, which is met by an edge running up to it from (4, 0), where two faces meet at right-angles. The hypoteneuse of the end right-angled triangle (in the y-z plane) is four units long and since it is isosceles, taking this as the base, the midpoint of it is below the apex, the line between them forming two similar (and therefore isosceles) right-angle triangles inside and "back to back", so the height must be two units. Using the co-ordinates of its vertices, we can work out the area of this triangle, by treating the edge which follows part of the y-axis as our "base". We have the height as 4 units and the length of the base as 4 units. So the base area is 8 square units. Hence the volume of this second solid is 8*2/3, which is 16/3 cubic units, a quarter of the volume of the previous solid.