Use differentials to approximate the volume of the shell.The thickness of the shell is a = 0.5 centimeter. Use differentials to approximate the volume of the shell. ?? cm3 http://www.webassign.net/larcalc9/3-9-036-alt.gif

You need to use the formula that helps you to find the volume of sphere such that: `V = (4/3)pi*r^3` Notice that the problem provides the radius of sphere, hence, you need to substitute 100 for r such that: `V =(4/3)pi*100^3` You need to differentiate both sides such that: `dV = ((4/3)pi*r^3) dr => dV = (4/3)*pi*3r^2 dr` Reducing by 3 yields: `dV = 4pi*r^2 dr` Substituting the `0.5` for `dr` yields: `dV = 4pi*100^2*0.5 => dV = 4pi*100*50 => dV = 20000pi~~ 62831.35 cm^3` Hence, evaluating the volume of the shell, under the given conditions, yields `dV =~~ 62831.35 cm^3` .

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Use differentials to approximate the volume of the shell.The thickness of the shell is a = 0.5 centimeter. Use differentials to approximate the volume of the shell. ?? cm3...

Use differentials to approximate the volume of the shell.

The thickness of the shell is a = 0.5 centimeter. Use differentials to approximate the volume of the shell.

?? cm3

http://www.webassign.net/larcalc9/3-9-036-alt.gif

Expert Answers

sciencesolve

| Certified Educator

You need to use the formula that helps you to find the volume of sphere such that:

`V = (4/3)pi*r^3`

Notice that the problem provides the radius of sphere, hence, you need to substitute 100 for r such that:

`V =(4/3)pi*100^3`

You need to differentiate both sides such that:

`dV = ((4/3)pi*r^3) dr => dV = (4/3)*pi*3r^2 dr`

Reducing by 3 yields:

`dV = 4pi*r^2 dr`

Substituting the `0.5` for `dr` yields:

`dV = 4pi*100^2*0.5 => dV = 4pi*100*50 => dV = 20000pi~~ 62831.35 cm^3`

Hence, evaluating the volume of the shell, under the given conditions, yields `dV =~~ 62831.35 cm^3` .

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Use differentials to approximate the volume of the shell.The thickness of the shell is a = 0.5 centimeter. Use differentials to approximate the volume of the shell. ?? cm3...

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