A particular tank is a sphere of radius 6 meters. Fluid to a depth of half the radius is in the tank. The density of the fluid is given by 8(h) = (16 – h2) kg/m3 at a height h meters up from the bottom of the tank (so h = 0 is the bottom of the tank and h = 12 is the top of the tank). The total mass of the fluid can be written as an integral of
the form S f(h)dh. (a) What is the lower limit of integration, a? (b) What is the upper limit of integration, b? (c)
What is the integrand, f(h)? (d) The integral gives the total mass of the fluid. What are the units of mass here? (Enter a whole plural word.)
The Correct Answer and Explanation is:
We are given a spherical tank of radius 6 meters, partially filled with fluid to a depth of half the radius. The density of the fluid varies with height hh (from the bottom of the tank) and is given by: ρ(h)=16−h2(in kg/m3)\rho(h) = 16 – h^2 \quad \text{(in kg/m}^3\text{)}
We are asked to express the mass of the fluid using an integral of the form: ∫abf(h) dh\int_a^b f(h) \, dh
(a) Lower limit of integration, aa:
Since height is measured from the bottom of the tank, the fluid starts at h=0h = 0.
✅ Answer: 0
(b) Upper limit of integration, bb:
The depth of the fluid is half the radius, so it fills up to: 62=3 meters from the bottom\frac{6}{2} = 3 \text{ meters from the bottom}
✅ Answer: 3
(c) Integrand, f(h)f(h):
To find total mass, we integrate density × volume element over height hh.
At height hh, a horizontal slice of the fluid is a circular disk. The volume element is: dV=πr(h)2 dhdV = \pi r(h)^2 \, dh
We need to find r(h)r(h), the radius of a horizontal slice at height hh, using geometry of the sphere.
The center of the sphere is at h=6h = 6. So the vertical distance from the center to height hh is ∣h−6∣|h – 6|. By the Pythagorean Theorem: r(h)2=62−(h−6)2=36−(h−6)2r(h)^2 = 6^2 – (h – 6)^2 = 36 – (h – 6)^2
So, f(h)=ρ(h)⋅area of slice=(16−h2)⋅π[36−(h−6)2]f(h) = \rho(h) \cdot \text{area of slice} = (16 – h^2) \cdot \pi [36 – (h – 6)^2]
✅ Answer: f(h)=(16−h2)⋅π[36−(h−6)2]f(h) = (16 – h^2) \cdot \pi [36 – (h – 6)^2]
(d) Units of mass:
Density is in kg/m³, volume in m³, so mass is: kg/m3×m3=kg\text{kg/m}^3 \times \text{m}^3 = \text{kg}
✅ Answer: kilograms
This problem involves determining the mass of a variable-density fluid inside a spherical tank using calculus. Since the tank is spherical with a radius of 6 meters and filled to half that radius (3 meters), we need to integrate over the vertical section of fluid from height 0 (bottom) to 3 meters (fluid surface). This sets the lower and upper limits of the integral: a=0a = 0 and b=3b = 3.
To compute the total mass, we use the principle: Mass=∫density×volume element\text{Mass} = \int \text{density} \times \text{volume element}
At each height hh, the fluid forms a thin horizontal disk. Its volume is dV=πr(h)2 dhdV = \pi r(h)^2 \, dh, where r(h)r(h) is the radius of the disk at height hh. Using geometry of a sphere centered at height 6 (since the radius is 6 m), the horizontal radius at height hh is: r(h)2=36−(h−6)2r(h)^2 = 36 – (h – 6)^2
The density varies with height as ρ(h)=16−h2\rho(h) = 16 – h^2. Thus, the integrand f(h)f(h) becomes: f(h)=ρ(h)⋅πr(h)2=(16−h2)⋅π[36−(h−6)2]f(h) = \rho(h) \cdot \pi r(h)^2 = (16 – h^2) \cdot \pi [36 – (h – 6)^2]
Multiplying density and cross-sectional area at each level gives mass of that slice. Integrating from 0 to 3 meters gives the total mass of fluid. Since density is in kg/m³ and volume in m³, the resulting units are kilograms, which is the standard unit of mass in the SI system.
