How can I complete this related rates math problem?
#Calculus Related Rates Problem
stiff jacinth
bright lavaBOT
quick spindle
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we need to get a function of the volume of a sphere whose input is h. We need to write volume in terms of the varying varying h.
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to do this let’s use integration. Remember how all an integral is, conceptually, is the sum of manyyyyy rectangles that are superrrrrrr tiny? Stack those oddly shaped rectangle together, and they look a lot like areas under cosine and x^3 and so on.
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In this case let’s take the sum of manyyyyy discs that are superrrr small, so when you stack them they form our sphere, the bowl of the fish!
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To form a volume of a partially filled sphere, let’s start by describing the integral of dV better. Let’s randomly take a realllyyyyyyy small disc out of our sphere. It has finite area pi r(h)^2, where r(h) a function of h. Note this r(h) radius of each flat disc and not the R radius of the whole 3D sphere. And this disc has verrrryy small height, dh. So dV = pi r(h)^2 dh. So our integral can be rewritten as “the integral of pi r(h)^2 dh”, or “the sum of manyyy discs of smalllll thickness”
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Side quest: To find r(h), let’s think of h as the x axis and r as the y axis. If you look at our sphere of radius R from the side, it’s a circle, duh! And this circle is represented by r(h)^2 + (h-R)^2 = R^2. Remember r(h) is the radius of each flat disc, parallel to the xy plane. Solving for r(h)^2 we get R^2 - (h-R)^3.
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Main quest: back to our integral. Plugging in for r(h)^2, we get “integral of pi ( R^2 - (h-R)^2 ) dh”. The bounds of the integral are the bounds of the integration variable h height. And height varies from zero and the actual height H. Now we can finally integrate this to get our formula for V(H)! It should have a RH^2 and a H^3.