I am stuck because I am not sure how to draw the digram since it is below the x-

Question

I am stuck because I am not sure how to draw the digram since it is below the x-axis.

1.A vessel has the shape obtained by revolving about the y-axis the part of the parabola y=2((x^2)-4) lying below the x-axis. If x and y are in feet, how much work is required to pump all the water in the full vessel to a point 4 ft above its top? use the denstiy of water 62.4lb/ft^2 to calculate.

2. A solid generated by revolving the region bounded by y=sqrt(9-x^2) and y=0 about the y-axis. A hole, centered along the axis of revolution, is drilled through this solid so that one third of the volume is removed. Find the diameter of the hole.

Explanation / Answer

We need to use the disk method here, where we look at the work required to lift
circular disks of water of height dy and radius x a distance (4 - y) feet, where
we have -y since the parabola lies below the x-axis. As the centers of the disks
go from (0,-8) to (0,0) the radius of the disks goes from 0 to 2.

With y = 2*(x^2 - 4) we have x^2 = (y/2) + 4. The work done lifting a disk of height
dy and radius x a distance of (4 - y) feet is mgh = p*dV*g*(4 - y) where p is the
density of water, g is acceleration due to gravity and dV is the volume of the disk
at 'level' y.

Now dV = pi*(x^2)*dy = pi*((y/2) + 4) dy, so the integral we need to evaluate is

integral(from y=-8 to 0)(p*g*pi*((y/2) + 4)*(4 - y) dy) =

(1/2)*p*g*pi*integral(y=-8 to 0)((y + 8)*(4 - y) dy) =

(1/2)*p*g*pi*integral(y=-8 to 0)((32 - 4y - y^2) dy) =

(1/2)*p*g*pi*[32*y - 2*y^2 - (1/3)*y^3] (from y = -8 to y = 0) =

(1/2)*p*g*pi*[0 - (32*(-8) - 2*(-8)^2 - (1/3)*(-8)^3)] =

(1/2)*p*g*pi*(640/3) = (320/3)*p*g*pi.

Now p = 62.4 pounds/ft.^3 and g = 32 feet/s^2 so the work done comes out to

669143.1 (pounds*ft^2/s^2). (I believe the units are called foot-pounds,

but I gave the precise units involved to be sure.)

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