By the fundamental theorem of calculus,
Now the arc length over an arbitrary interval
But before we compute the integral, first we need to make sure the integrand exists over it.
I don’t understand how to solve this : -17 + 18 - 12
2 answers:
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By the fundamental theorem of calculus,
Now the arc length over an arbitrary interval
But before we compute the integral, first we need to make sure the integrand exists over it.
So hmm check the picture below
the radius "r" is half the diameter, meaning, the diameter is 2r long
now, if the height "h" is twice "d" or 2d, then that means h = 2(2r)
thus![\bf \textit{volume of a cone}\\\\ V=\cfrac{\pi r^2 h}{3}\qquad \begin{cases} h=height\\ r=radius\\ ------\\ h=2(2r) \end{cases}\implies V=\cfrac{\pi r^2[2(2r)]}{3} \\\\\\ V=\cfrac{4\pi r^3}{3}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bvolume%20of%20a%20cone%7D%5C%5C%5C%5C%0AV%3D%5Ccfrac%7B%5Cpi%20r%5E2%20h%7D%7B3%7D%5Cqquad%20%0A%5Cbegin%7Bcases%7D%0Ah%3Dheight%5C%5C%0Ar%3Dradius%5C%5C%0A------%5C%5C%0Ah%3D2%282r%29%0A%5Cend%7Bcases%7D%5Cimplies%20V%3D%5Ccfrac%7B%5Cpi%20r%5E2%5B2%282r%29%5D%7D%7B3%7D%0A%5C%5C%5C%5C%5C%5C%0AV%3D%5Ccfrac%7B4%5Cpi%20r%5E3%7D%7B3%7D)
the radius "r" is half the diameter, meaning, the diameter is 2r long
now, if the height "h" is twice "d" or 2d, then that means h = 2(2r)
thus