A sphere of radius 3 feet contains grain with a depth of 5 feet. The grain inside weighs 5 pounds per cubic foot.
Find the amount of work in foot-pounds required to empty the trough by lifting it to the top of the sphere.
AAAaannndd the bot gets it wrong yet again!
at a depth of 3-h, the radius of the cross-section is r^2 = 3^2-h^2
so the work per slice of thickness dh is weight * distance
∫[-3,2] 5π(9-h^2)(3-h) dh = 2125π/4 = 1669
Thank you!
how did you get the integral with [-3,2]?
To find the amount of work required to empty the trough by lifting it to the top of the sphere, we need to calculate the volume of the grain and then multiply it by the weight of the grain.
Step 1: Calculate the volume of the grain.
The trough is in the shape of a sphere with a radius of 3 feet. The grain inside has a depth of 5 feet. We can find the volume of the grain by subtracting the volume of the empty cavity from the volume of the sphere.
The volume of the sphere can be calculated using the formula:
V = (4/3) * π * r^3
where r is the radius of the sphere.
Given r = 3,
V = (4/3) * π * 3^3
V = (4/3) * π * 27
V = 36π
The cavity inside the sphere represents a cone, and its volume can be calculated using the formula:
V = (1/3) * π * r^2 * h
where r is the radius of the base and h is the height (or depth) of the cone.
Given r = 3 and h = 5,
V_cone = (1/3) * π * 3^2 * 5
V_cone = (1/3) * π * 9 * 5
V_cone = 15π
So, the volume of the grain is the difference between the volume of the sphere and the volume of the cavity inside the sphere:
V_grain = V - V_cone
V_grain = 36π - 15π
V_grain = 21π
Step 2: Calculate the work required.
The grain weighs 5 pounds per cubic foot, which means for every 1 cubic foot of grain, the work required to lift it is 5 foot-pounds.
We have already calculated the volume of the grain as V_grain = 21π cubic feet.
Therefore, the work required to lift the grain to the top of the sphere is given by:
Work = 5 * V_grain
Work = 5 * 21π
Work = 105π
So, the amount of work required to empty the trough by lifting it to the top of the sphere is 105π foot-pounds.