Find the volume of the solid obtained by rotating the region

Solution

1. This little \"triangular\" region can be subdivided into horizontal rectangles, so I will use disks to find the volume. dV = px^2 dy = p (7y^2)^2 dy {from y = 0 to y = 1} V = 49p ? y^4 dy = 49p {y^5/5} [0,1] = 49p (1/5 - 0) = 49p/5 2. Now we have to take into account that the axis of revolution is not an axis, but y = 6, which is horizontal. So washers are used to find the volume. dV = p[(6 - y1)^2 - (6 - y2)^2] dx = p{[6- (vx)]^2 - [6 -(x/6)]^2} dx] V = 82.42449214...p ˜ 258.944...