Method of Cylindrical Shells (Part 3)

[shamieh]

The region bounded by the given curves is rotated about the specified axis. Find the volume of the resulting solid by any method..
\(\displaystyle y = -x^2 + 6x - 8, y = 0\)

so I got -8 to 0 for the integral by plotting the graph... How are they getting 2 to 4? You can't solve that by factoring? And when i plugged it into the quadratic formula i got 1 to 5

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The region bounded by the given curves is rotated about the specified axis. Find the volume of the resulting solid by any method..
\(\displaystyle y = -x^2 + 6x - 8, y = 0\)

so I got -8 to 0 for the integral by plotting the graph... How are they getting 2 to 4? You can't solve that by factoring? And when i plugged it into the quadratic formula i got 1 to 5

nevermind my god I am stupid I forgot to distribute the x

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Actually I take that back, I still don't understand how they get 2 to 4.

 

[MarkFL]

\(\displaystyle y=-x^2+6x-8=-\left(x^2-6x+8 \right)=-(x-2)(x-4)\)

What is the axis of rotation?

 

[shamieh]

Oh i was the - sign... I see(Worried)

 

[shamieh]

\(\displaystyle y=-x^2+6x-8=-\left(x^2-6x+8 \right)=-(x-2)(x-4)\)

What is the axis of rotation?

x axis?

 

[MarkFL]

x axis?

Which method do you feel will be more straightforward, and can you give an element of the volume, that is, the volume of an arbitrary shell, disk or washer (depending on the method you choose)?

I actually recommend doing these problems more than one way when possible just for the practice and as a means of checking your result.

 

[MarkFL]

We can work this type of problem in general by requiring one of the roots of the parabolic boundary to be the origin, and the other on the positive $x$-axis.

So let the region be bounded by:

\(\displaystyle y=-kx(x-a)\) and \(\displaystyle y=0\), which is also the axis of rotation.

Note: \(\displaystyle 0<k\) and \(\displaystyle 0<a\).

Disk method:

The volume of an arbitrary disk is:

\(\displaystyle dV=\pi r^2\,dx\)

where:

\(\displaystyle r=y=-kx(x-a)\)

hence:

\(\displaystyle dV=\pi \left(-kx(x-a) \right)^2\,dx=\pi k^2\left(x^4-2ax^3+a^2x^2 \right)\,dx\)

Summing up the disks, we find:

\(\displaystyle V=\pi k^2\int_0^a x^4-2ax^3+a^2x^2\,dx\)

Applying the FTOC, we obtain:

\(\displaystyle V=\pi k^2\left[\frac{1}{5}x^5-\frac{a}{2}x^4+\frac{a^2}{3}x^3 \right]_0^a=\pi a^5k^2\left(\frac{1}{5}-\frac{1}{2}+\frac{1}{3} \right)=\frac{\pi a^5k^2}{30}\)

Shell method:

The volume of an arbitrary shell is:

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

where:

\(\displaystyle r=y\)

To find $h$, consider:

\(\displaystyle y=-kx(x-a)=-kx^2+akx\)

Arranged in standard quadratic form:

\(\displaystyle kx^2-akx+y=0\)

Application of the quadratic formula yields:

\(\displaystyle x=\frac{ak\pm\sqrt{a^2k^2-4ky}}{2k}\)

Hence:

\(\displaystyle h=\frac{ak+\sqrt{a^2k^2-4ky}}{2k}-\frac{ak-\sqrt{a^2k^2-4ky}}{2k}=\frac{\sqrt{a^2k^2-4ky}}{k}\)

And so we have:

\(\displaystyle dV=\frac{2\pi}{k}\left(y\sqrt{a^2k^2-4ky} \right)\,dy\)

To determine the upper limit of integration, we note that the axis of symmetry for the parabola is on \(\displaystyle x=\frac{a}{2}\) and so the upper limit is:

\(\displaystyle y=-k\frac{a}{2}\left(\frac{a}{2}-a \right)=\frac{a^2k}{4}\)

And so the volume is given by:

\(\displaystyle V=\frac{2\pi}{k}\int_0^{\frac{a^2k}{4}} y\sqrt{a^2k^2-4ky}\,dy\)

At this point, we may want to develop a formula for the indefinite integral:

\(\displaystyle I=\int x\sqrt{a+bx}\,dx\)

Let's try the substitution:

\(\displaystyle u=a+bx\,\therefore\,du=b\,dx\)

Hence, the integral becomes:

\(\displaystyle I=\frac{1}{b^2}\int u^{\frac{3}{2}}-au^{\frac{1}{2}}\,du\)

Using the power rule, we find:

\(\displaystyle I=\frac{1}{b^2}\left(\frac{2}{5}u^{\frac{5}{2}}-\frac{2a}{3}u^{\frac{3}{2}} \right)+C=\frac{2}{15b^2}u^{\frac{3}{2}}\left(3u-5a \right)+C\)

Back-substituting for $u$, we obtain:

\(\displaystyle I=\frac{2}{15b^2}(a+bx)^{\frac{3}{2}}\left(3(a+bx)-5a \right)+C\)

\(\displaystyle I=\frac{2}{15b^2}(a+bx)^{\frac{3}{2}}\left(3bx-2a \right)+C\)

Now, applying this to our volume, we get through the application of the FTOC:

\(\displaystyle V=\frac{\pi}{60k^3}\left[\left(a^2k^2-4ky \right)^{\frac{3}{2}}(-12ky-2a^2k^2) \right]_0^{\frac{a^2k}{4}}=\frac{\pi}{60k^3}\left(0+2a^5k^5 \right)\)

And thus:

\(\displaystyle V=\frac{\pi a^5k^2}{30}\)

Applying this formula to the given problem, we then find:

\(\displaystyle V=\frac{\pi 2^51^2}{30}=\frac{16\pi}{15}\)