Problem Statement:
A bowl is a solid of revolution bounded by the surfaces obtained by rotating the curves
\[C_1: x^2=4y \quad \text{and} \quad C_2: x^2=8(y-k)\]
about the y-axis, where \(k\) is a constant greater than zero.
(a) What is the diameter of the mouth of the bowl in terms of \(k\)?
(b) Find the capacity of the bowl in terms of \(\pi\) and \(k\)
(c) Water is poured into the bowl so that its greatest depth is \(\frac{k}{2}\). What fraction of the capacity of the bowl is filled?
(d) Show that for all positive values of \(k\), the volume of the material used in making the bowl is equal to the capacity of the bowl
Answer:
(a)
To find the intersection between \(C_1\) and \( C_2\),
\[4y = 8(y-k)\]
\[4y = 8y-8k\]
\[y = 2k\]
Substitute in \(C_1\), \[x^2=4(2k)\]
\[x=\pm2\sqrt{2k}\]
∴ diameter = \(2\sqrt{2k}-(-2\sqrt{2k}) = 4\sqrt{2k}\)
(b)
For \(C_2\), the minimum point is \((0, k)\)
\[\text{Capacity of Bowl} = \pi \int_{k}^{2k} x^2 \, dy\]
\[V_B = \pi \int_{k}^{2k} 8(y-k) \, dy\]
\[V_B = \pi [4y^2-8ky]_{k}^{2k}\]
\[V_B = \pi [(16k^2-16k^2)-(4k^2-8k^2)]\]
\[V_B = 4k^2 \pi\]
(c)
\[\text{Capacity of bowl that is filled} = \pi \int_{k}^{k+\frac{k}{2}} 8(y-k) \, dy\]
\[V_f = \pi [4y^2-8ky]_{k}^{\frac{3k}{2}}\]
\[V_f = \pi [(9k^2-12k^2)-(4k^2-8k^2)]\]
\[V_f = k^2\pi\]
\[\text{Fraction of capacity of the bowl filled} = \frac{V_f}{V_B} = \frac{k^2\pi}{4k^2\pi} = \frac{1}{4}\]
(d)
\[\text{Volume of material used} = \pi \int_{0}^{2k} 4y \,dy \quad – \quad \pi \int_{k}^{2k} 8(y-k) \,dy\]
\[V_m = \pi [2y^2]_{0}^{2k} \quad – \quad \pi [4y^2-8ky]_{k}^{2k}\]
\[V_m = 8k^2 \pi \quad – \quad \pi[(16k^2-16k^2)-(4k^2-8k^2)]\]
\[V_m = 8k^2 \pi – 4k^2 \pi\]
\[V_m = 4k^2 \pi\]
\[V_m = V_B\]
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