Method of Cylindrical Shells Question #2

[shamieh]

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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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 ?

Nevermind i see what is going on... 2pi * X

so that's why you set x =

 

[MarkFL]

If you are to use the shell method, you want to observe that the volume of an arbitrary shell is:

\(\displaystyle dV=2\pi rh\,dy\)

where:

\(\displaystyle r=y\)

\(\displaystyle h=x=y^{\frac{1}{3}}\)

hence:

\(\displaystyle dV=2\pi y^{\frac{4}{3}}\,dy\)

Summing up all the shells, we find:

\(\displaystyle V=2\pi\int_0^8 y^{\frac{4}{3}}\,dy\)

If you wish to check your work by using the washer method, the volume of an arbitray washer is:

\(\displaystyle dV=\pi\left(R^2-r^2 \right)\,dx\)

where:

\(\displaystyle R=8\)

\(\displaystyle r=y=x^3\)

hence:

\(\displaystyle V=\pi\int_0^2 8^2-x^6\,dx\)

You should verify that both definite integrals give the same result.