Let R be the region bounded by yx2 and y 4 Compute the volu

Solution

I won\'t show sketches, but you should sketch it out. This is a fairly simple system. The region is a triangle with vertices at (0,0), (0,3), and (3,0). A and b are solid cones with the same volume. Volume of a cylinder is V = PI * r^2 * h Approximate the volume of all circular cylinder slice perpendicular to the base of the cone to be: V = PI * int(r^2, dh). a) r in this case would be x and dh is dy. From the region, x=3-y. V = PI * int((3-y)^2, 0, 3, dy) = PI * int( y^2-6y+9, 0, 3, dy) = PI * [(1/3)y^3-3y^2+9y, 0, 3] = 9 * PI = 28.274 b) just switch x and y from part a c and e are bowls with the same volume. The volume of a washer is calculated by V = PI * (ro^2 - ri^2) * h. ro is the outer radius and ri is the inner radius. The integral volume of all washer slices perpendicular to the base of the bowl would be V = PI * int(ro^2 - ri^2, dh). The inner radius is x = -y. The outer radius is x=-3. Remember, the new axis is around y=3, not the y axis, so the equations were adjusted by -3. V = PI * int((-y)^2 - (-3)^2), 0, 3, dy) = 18 * PI = 56.55 d and f are... umm... hollow tubes with an outer diameter that expands. Same washer idea. For d: V = PI * int(ro^2 - ri^2, dh) = PI * int((6-y)^2 - (-3)^2), 0, 3, dy) = 36 * PI = 113.10