Question

Drinking juice A glass has circular cross sections that taper (linearly) from a radius of 5 cm at the top of the glass to a radius of 4 cm at the bottom. The glass is 15 cm high and full of orange juice. How much work is required to drink all the juice through a straw if your mouth is 5 cm above the top of the glass? Assume the density of orange juice equals the density of water.

Accepted Answer

Identify the physical setup: The glass is a frustum of a cone with a height of 15 cm, where the radius decreases linearly from 5 cm at the top to 4 cm at the bottom. The juice must be lifted to a point 5 cm above the top of the glass, so the total lifting height varies with the depth of the juice. Set up a coordinate system: Let the vertical coordinate $$ x $$ measure the depth from the top of the glass downward, with $$ x=0  at $$the top and $$ x=15  at $$the bottom. The radius $$ r(x) $$ changes linearly from 5 cm at $$ x=0  to 4 cm at$$ $$ x=15 . $$Express $$ r(x)  as a $$linear function of $$ x . $$Express the volume of a thin horizontal slice of juice: Consider a thin slice of thickness $$ dx  at $$depth $$ x . $$The volume of this slice is $$ dV = \pi (r(x))^2 dx . $$Determine the work to lift this slice: The weight of the slice is $$ dW_{weight} = \rho g dV $$, where $$ \rho  is $$the density of the juice (same as water) and $$ g  is $$acceleration due to gravity. The distance this slice must be lifted is $$ (x + 5)  cm$$, since the juice at depth $$ x $$ must be raised to 5 cm above the top (which is at $$ x=0 ). $$Set up the integral for total work: The total work is the integral from $$ x=0  to$$ $$ x=15  of $$the weight of each slice times the lifting distance, i.e., $$ W = \int_0^{15} \rho g \pi (r(x))^2 (x + 5) \, dx . $$Substitute the expression for $$ r(x) $$ and constants, then evaluate the integral to find the total work.

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