differentials

kingaaron08041991

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use differentials to estimate the amount of paint needed to apply a coat of pain 0.05 cm thick to a hemispherical dome with daimater 50m
 
\(\displaystyle Common \ Sense \ Approach, \ no \ calculus.\)

\(\displaystyle V_{sphere} \ = \ \frac{4\pi r^{3}}{3}, \ V_{hemisphers} \ = \ \frac{2\pi r^{3}}{3}\)

\(\displaystyle D \ = \ 50m, \ r \ = \ 25m, \ .05cm \ = \ .0005m\)

\(\displaystyle Volume \ of \ hemi \ before \ painting: \ V \ = \ (2/3)\pi(25)^{3} \ = \ 32,724.9234749m^{3}\)

\(\displaystyle Volume \ of \ hemi \ after \ painting: \ V \ = \ (2/3)\pi(25.0005)^{3} \ = \ 32,726.8870096m^{3}\)

\(\displaystyle Difference \ in \ volume \ = \ 1.963534m^{3}, \ hence \ the \ painter \ will \ need \ about \ 2m^{3} \ of \ paint.\)



\(\displaystyle Calculus \ approach:\)

\(\displaystyle V_{hemisphere} \ = \ \frac{2\pi r^{3}}{3}, \ r \ = \ 25m, \ dr \ = \ .0005m\)

\(\displaystyle dV \ = \ 2\pi r^{2}dr \ = \ 2\pi(25)^{2}(.0005) \ = \ 1.9634954084m^{3} \ of \ paint \ needed.\)
 
kingaaron08041991 said:
use differentials to estimate the amount of paint needed to apply a coat of pain 0.05 cm thick to a hemispherical dome with daimater 50m

Since the instruction in the problem requires you to use differentials,

Area of hemisphere = A = ?/2 * D[sup:ydz4mi1k]2[/sup:ydz4mi1k]

?A = ? * D * ?D

?A = ? * 50 * 0.001 m[sup:ydz4mi1k]2[/sup:ydz4mi1k] = 0.05 * ? m[sup:ydz4mi1k]2[/sup:ydz4mi1k]

V = ?/12 * D[sup:ydz4mi1k]3[/sup:ydz4mi1k] = 1/6 * A * D

?V = 1/6 (A * ?D + D * ?A)

Now finish it .....

You can check the answer from Glen's solution.
 
Another way (more direct):

V = ?/12 * D[sup:2tvxho5n]3[/sup:2tvxho5n]

?V = ?/4 * D[sup:2tvxho5n]2[/sup:2tvxho5n] * ?D = ?/4 * (50)[sup:2tvxho5n]2[/sup:2tvxho5n] * 0.001 m[sup:2tvxho5n]3[/sup:2tvxho5n]
 
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