MHB 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...$$y = x^3$$ , $$y = 8$$ and $$x = 0$$

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 $$2\pi \int^8_1 x(x^2)$$ 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...$$y = x^3$$ , $$y = 8$$ and $$x = 0$$

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 $$2\pi \int^8_1 x(x^2)$$ 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:

$$dV=2\pi rh\,dy$$

where:

$$r=y$$

$$h=x=y^{\frac{1}{3}}$$

hence:

$$dV=2\pi y^{\frac{4}{3}}\,dy$$

Summing up all the shells, we find:

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

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

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

where:

$$R=8$$

$$r=y=x^3$$

hence:

$$V=\pi\int_0^2 8^2-x^6\,dx$$

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