Expert: Scotto - 4/15/2010

Question
1. Find the area of the region bounded by the curve y2 = x and the line x - 4=0
2. fin the volume of the solid generated by revolving about the line x = 2 and the area bounded by y = x2, x = 2 and y = o.
3. the natural length of a spring is 8 in. a force of 20lb stretches the spring to 10 in. Find the work done in stretching the spring form its natural lenght to 12 in.
4. a conical tank is 12ft across the top and 8ft deep. It conatins water to depth of 6ft. Find the work done in pumping the water to a point 2 ft above the top of the tank.
5. the outlet of a reservoir closes a circular hole in the side. Find the force on the outlet gate if the hole is 4 ft, in diameter and its center is 40ft below the water level.

Answer
1. ∫y²dy from 2 downto -2.  Note ∫y^n dy = y^(n+1) / (n+1).

2. The integral will be found by rotating x² around x = 2.
That means the radius will be 2 - x² for y going fron 0 to 4.
Since we need y, we know y = 2 - x², so x =  √(2-y).
Since that is the radius, the area at any y is πx², and since x = √(2-y), that is π(2-y).

To get the answer, integerate that from 0 to 4.

3. It is known that work W = Fd where F is the force and d is the distance.
Since it requires 20 lb to stretch it 2 inches (from 8 to 10),
it requires 20 lb more to stretch it 2 more inches.
This is a total of 40 lb over 4 inches.

Remember, F = Wd, that is, force = work * distance.

4. Taking 2 inces above the tank as 0, this means the water is lifted from -2' to -8'.
It also means the amount of water at each level is πr², were r is x².
Now at x = -8, the radius r is 0' and at height -2 to radius is 3/4 of the radius at the top.
This means it as a (3/4)12' = 9' radius at that depth.  This means the radius is 3/2 the width.

This gives us the equation ∫π(3(x+8)/2)²dx from x = -8 down to x = -2.

5. The force is due to the depth times the width of the hole at that depth.
Since the diameter is 4, the radius is 2.
The equation of the circle is x² + (y+40)² = 4.
At depth y, the width is then x = ±√(4-(y+40)²).
The solution is then found in ∫2√(4-(y+40)²)dy, y going from -38 down to -40.