Rates of change: inflated hot-air balloon

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Homework Statement


An inflated hot air balloon can be considered as a perfect sphere woth diameter of 30m. When the hatch is opened, the hot air is lost at a rate of 118m^3/s. For the first 5 seconds, the balloon maintains its spherical shape, but then it begins to collapse. At what rate is the diameter decreasing after 3 seconds? How long does the balloon take to completely collapse?

I know this is a chain rule question where dV/dt=118m^3/s and I'm not sure how to relate that to the diameter? I know that, because dV/dt is part of the equation the equation is either dV/dt=dV/dD*dD/dt where D is the diameter or the equation dD/dt=dD/dV*dV/dt or am I going about this the completely wrong way and have to get everything in terms of the radius?
 
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do you know the formula for the volume of a sphere in terms of the radius?
 
4/3∏r^3
 
Then, what is the volume of a sphere in terms of its diameter?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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