Nick's question at Yahoo Answers regarding a volume by slicing

[MarkFL]

Here is the question:

Finding volume of a solid (calculus)?

Find the volume of the solid with the given base and cross sections.

The base is the triangle enclosed by x + y = 9,
the x-axis, and the y-axis. The cross sections perpendicular to the y-axis are semicircles.

Thanks in advance.

I have posted a link there to this thread so the OP can view my work.

 

[MarkFL]

Hello Nick,

For any slice of the solid made perpendicular to the $y$-axis, the diameter of the semi-circle is the $x$-coordinate on the line $x+y=9$, and so the radius is:

\(\displaystyle r=\frac{x}{2}=\frac{9-y}{2}\)

And so the volume of an arbitrary slice is:

\(\displaystyle dV=\pi r^2\,dy=\frac{\pi}{4}(9-y)^2\,dy\)

Hence, the summation of all the slices is given by:

\(\displaystyle V=\frac{\pi}{4}\int_0^9 (9-y)^2\,dy\)

Using the substitution:

\(\displaystyle u=9-y\,\therefore\,du=-dy\) we obtain:

\(\displaystyle V=-\frac{\pi}{4}\int_9^0 u^2\,dy\)

Using the rule:

\(\displaystyle -\int_a^b f(x)\,dx=\int_b^a f(x)\,dx\) we may write:

\(\displaystyle V=\frac{\pi}{4}\int_0^9 u^2\,dy\)

Applying the FTOC, we obtain:

\(\displaystyle V=\frac{\pi}{12}\left[u^3 \right]_0^9=\frac{9^3\pi}{12}=\frac{243\pi}{4}\)