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Use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by the given curves about the x axis...
\(\displaystyle y = x^3\) , \(\displaystyle y = 8\) and \(\displaystyle x = 0\)
So my question is: Why did they cube root the y (to be more technical why did they put it in terms of x??? I don't understand what this is accomplishing? Can't you just set up your graph and have a horizontal asymptote at y = 8, a parabola that doesn't pass (2,8), and then just set up your integral and solve as \(\displaystyle 2\pi \int^8_1 x(x^2)\) dx ???
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so thats why you set x =
\(\displaystyle y = x^3\) , \(\displaystyle y = 8\) and \(\displaystyle x = 0\)
So my question is: Why did they cube root the y (to be more technical why did they put it in terms of x??? I don't understand what this is accomplishing? Can't you just set up your graph and have a horizontal asymptote at y = 8, a parabola that doesn't pass (2,8), and then just set up your integral and solve as \(\displaystyle 2\pi \int^8_1 x(x^2)\) dx ???
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Nevermind i see what is going on... 2pi * XUse the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by the given curves about the x axis...
\(\displaystyle y = x^3\) , \(\displaystyle y = 8\) and \(\displaystyle x = 0\)
So my question is: Why did they cube root the y (to be more technical why did they put it in terms of x??? I don't understand what this is accomplishing? Can't you just set up your graph and have a horizontal asymptote at y = 8, a parabola that doesn't pass (2,8), and then just set up your integral and solve as \(\displaystyle 2\pi \int^8_1 x(x^2)\) dx ???
so thats why you set x =