Part b: setting up integrals
a circular ring is made of a material with uniform density p (kg/m²).
the ring has rectangular cross section, as shown in the figure to the
right. the outer radius of the ring is r.
we would like to calculate the total mass of the ring.
first we will use the approximation that the width of the ring, a, is very much smaller than
the outer radius of the ring r(a < < r).
a) on the figure to the right, label a small "chunk"
of the ring that subtends an arc de.
b) what is the mass of this chunk (dm) in terms of
p. a, b, r, and de?
c) express the mass of the entire ring as a single integral in the angle 0. (don't forget limits)
d) what is the total mass of this ring?

b) dm = ρab R dθ

c) m = ∫ ρab R dθ, from θ = 0 to θ = 2π

d) m = 2πρabR

Explanation:

b) We want to find the mass dm of a small chunk of the ring.

Mass is density times volume:

dm = ρ dV

Since a << R, we can approximate the volume as a rolled rectangular prism.  Therefore, the volume of the chunk is the area of the cross section times the arc length.

dV = ab R dθ

dm = ρab R dθ

c) The mass of the entire ring is the sum of the masses of all the chunks.

m = ∫ dm

m = ∫ ρab R dθ, from θ = 0 to θ = 2π

d) ρ, a, b, and R are constants, so:

m = ρabR ∫ dθ

Evaluating the integral:

m = ρabR (θ|0 to 2π)

m = ρabR (2π − 0)

m = 2πρabR

the temperature in kelvin of 2.0 dm3 of an ideal gas is doubled and its pressure is increased by a factor of four. what is the final volume of the gas

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