MHB *est amt of paint to a coat of paint 0.05 cm thick

[karush]

Use differentials to estimate the amount of paint needed
to apply a coat of paint $$0.05 \text{ cm}$$ thick
to a hemispherical dome with a diameter of $$50\text{ m}$$
$\displaystyle V_h=\frac{1}{2} \cdot\frac{4}{3}\pi\text{r}^3 = \frac{4}{6}\pi\text{r}^3$
$dV_h = 2\pi\cdot r^2\cdot \text{dr}$
so if $$r=25\text{ m} = 2500\text{ cm}\text { and }dr = 0.05\text{ cm}$$ then
$dV_h = 2\pi \ 2500\text{ cm}^2\cdot 0.05\text{ cm}\approx 1.96\text{ m}^3$
or should $r=2500.05\text{ cm}$

 

[MarkFL]

You did it correctly. I always like to, if possible, compare the approximate to the true value:

$$V=\frac{2\pi}{3}\left((r+\Delta r)^3-r^3 \right)$$

$$V=\frac{2\pi}{3}\left(2500.05^3-2500^3 \right)\text{ cm}^3\approx1.96353467866359\text{ m}^3$$

This is very close to the estimate.

 

[DreamWeaver]

(Rock)

Great thread! lol[I'm a painter and decorator, see, so I might just find a use for this... (Hug) ]

 

[karush]

well a hemisphere would not be easy to paint especially .05 cm uniformly!

with a big brush I guess

 

[DreamWeaver]

well a hemisphere would not be easy to paint especially .05 cm uniformly!

with a big brush I guess

Only one way to get such even coverage... Mr Bean has the answer: Mr Bean - Painting with Fireworks - YouTube

;)