Volume by Slicing

[StintedVisions]

The problem is "Use the disk method to find the volume of the solid generated by revolving the shaded region shown to the right by the x-axis." And it has a picture with a graph with a right triangle with vertices at (0,0), (0,10), (6,0) and the equation 5x+3y=30 and a red line dissecting the triangle at x=2.

So I'm chugging along with the "help me solve this button" and get to this /pi$\int_0^6 ($\frac{30-5x}{32}$)^2~dx$

and it says it is
= 25/pi$\int_0^6 ($\2-frac{x}{3}$)^2~dx$

How are they pulling 25 out of that? I'm not quite seeing it. Maybe I just need to take a break from it.

 

[MarkFL]

The red line is the radius of an arbitrary disk. The given line is that along which the hypotenuse lies, and its value is that of the radii of the disks. The volume of an arbitrary disk is:

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

where:

\(\displaystyle r=y=\dfrac{30-5x}{3}=\dfrac{5(6-x)}{3}\)

and so we have:

\(\displaystyle dV=\pi\left(\dfrac{5(6-x)}{3} \right)^2\,dx=\pi\left(\dfrac{5}{3} \right)^2(6-x)^2\,dx=\dfrac{25\pi}{9}(6-x)^2\,dx\)

Summing all the disks, we find:

\(\displaystyle \displaystyle V=\frac{25\pi}{9}\int_0^6(6-x)^2\,dx\)

Does this make sense?

Now, rather than expanding the integrand, use the substitution:

\(\displaystyle u=6-x\,\therefore\,du=-dx\)

to get:

\(\displaystyle \displaystyle V=-\frac{25\pi}{9}\int_6^0 u^2\,du=\frac{25\pi}{9}\int_0^6 u^2\,du\)