Question

At a party, the punch bowl is hemispherical, with 2 foot radius and open at the top. The fruit punch weighs 66 pounds per cubic foot. At the beginning of the party the bowl is full to the brim. The rules for serving punch require each person to fill the ladle, raise it exactly 1 foot above the top of the owl (without spilling!), and then pour the punch into a cup at that level. After 10 minutes, 30 people have filled 30 cups, and the punch remaining in the bowl is 4 inches deep. Draw a picture of the situation, and then set up the integral for the work W involved in raising the punch until the punch remaining in the bowl has a depth of only 4 inches. Include all pertinent information in the integral, but do not evaluate the integral.

Answer 1

W = ∫ [−2/3 to 0] 66π (2−y)(1−y²) dy = 2332π/27

Explanation:

Vertical cross-section of bowl is a semi-circle of radius 1, centered at (0,0)

x² + y² = 1 (for y ≤ 0)

Bottom of bowl is at y = −1

Top of bowl is at y = 0

When punch is 4 inches deep, this 1/3 ft above bottom of bowl:

y = −1+1/3 = −2/3

So we integrate from y = −2/3 to y = 0

Radius of cross section at height of y is = x = √(1−y²)

Area of cross section at height y is πr²= πx² = π(1−y²)

Since each person is to lift ladle 2 feet above the top of the bowl, then distance = 2−y

W = Σ (density of punch * distance * volume of punch)

W = ∫ [−2/3 to 0] 66 * (2−y) * πr² dy

W = ∫ [−2/3 to 0] 66π (2−y)(1−y²) dy = 2332π/27