# Method of Cylindrical Shells Question #2

#### shamieh

##### Active member
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 thats why you set x =

#### MarkFL

Staff member
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.