HelloWorld's questions at Yahoo Questions invovling solids of revolution

[MarkFL]

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]

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:

View attachment 1334

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).