# HelloWorld's questions at Yahoo! Questions invovling solids of revolution

#### MarkFL

Staff member
Here are the questions:

Volume by shells method?

Find the volume by shells of the region bounded by:

x = y^2
y = x^2

a) Rotated about the x-axis

b) Rotated about the y-axis

How would you go about solving?
I have posted a link there to this topic so the OP can see my work.

#### MarkFL

Staff member
Hello HelloWorld,

First, we should observe that since there is cyclic symmetry between the variables, we should expect to get the same result revolving about either axis, however, I will work through both parts of the problem as if we do not know this. Also, I will use both the shell and washer methods, to demonstrate both methods, and as a check.

First, let's draw a diagram of the region to be revolved:

a) Revolving about the $x$-axis, using the shell method, we find the volume of an arbitrary shell is:

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

where:

$$\displaystyle r=y$$

$$\displaystyle h=\sqrt{y}-y^2=y^{\frac{1}{2}}-y^2$$

And so we have:

$$\displaystyle dV=2\pi y\left(y^{\frac{1}{2}}-y^2 \right)\,dy=2\pi \left(y^{\frac{3}{2}}-y^3 \right)\,dy$$

Summing the shells by integrating, we first determine the limits of integration. If we square the equation $x=y^2$ to get $x^2=y^4$, then equate the two expressions for $x^2$, we find:

$$\displaystyle y=y^4$$

$$\displaystyle y\left(1-y^3 \right)=0$$

$$\displaystyle y(1-y)\left(1+y+y^2 \right)=0$$

The two real roots are:

$$\displaystyle y=0,\,1$$

Which means the two curves intersect at the points $(0,0),\,(1,1)$.

Hence, we may find the volume with:

$$\displaystyle V=2\pi\int_0^1 y^{\frac{3}{2}}-y^3\,dy$$

Applying the FTOC, we have:

$$\displaystyle V=2\pi\left[\frac{2}{5}y^{\frac{5}{2}}-\frac{1}{4}y^4 \right]_0^1=2\pi\left(\frac{2}{5}-\frac{1}{4} \right)=\frac{3\pi}{10}$$

Now, let's check this by using the washer method. The volume of an arbitrary washer is:

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

where:

$$\displaystyle R=\sqrt{x}$$

$$\displaystyle r=x^2$$

and so we have:

$$\displaystyle dV=\pi\left(\left(\sqrt{x} \right)^2-\left(x^2 \right)^2 \right)\,dx=\pi\left(x-x^4 \right)\,dx$$

Summing the washers, we find:

$$\displaystyle V=pi\int_0^1 x-x^4\,dx$$

Applying the FTOC, we find:

$$\displaystyle V=\pi\left[\frac{1}{2}x^2-\frac{1}{5}x^5 \right]_0^1=\pi\left(\frac{1}{2}-\frac{1}{5} \right)=\frac{3\pi}{10}$$

And this checks with the result from the shell method.

b) Revolving about the $y$-axis, using the shell method, we find the volume of an arbitrary shell is:

$$\displaystyle dV=2\pi rh\,dx$$

where:

$$\displaystyle r=x$$

$$\displaystyle h=\sqrt{x}-x^2=x^{\frac{1}{2}}-x^2$$

And so we have:

$$\displaystyle dV=2\pi x\left(x^{\frac{1}{2}}-x^2 \right)\,dy=2\pi \left(x^{\frac{3}{2}}-x^3 \right)\,dx$$

Since we will integrate from $x=0$ to $x=1$ it now becomes obvious that we will get the same volume as in part a).