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

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The discussion focuses on calculating the volume of a solid with a triangular base defined by the line x + y = 9 and the axes, with semicircular cross sections perpendicular to the y-axis. The radius of the semicircle is derived from the x-coordinate, leading to the volume of an arbitrary slice expressed as dV = πr² dy. The total volume is computed using integration, resulting in V = (π/12)(9^3) = 243π/4. The solution involves substitution and the Fundamental Theorem of Calculus to evaluate the integral. The final volume of the solid is confirmed as 243π/4.
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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.
 
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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:

$$r=\frac{x}{2}=\frac{9-y}{2}$$

And so the volume of an arbitrary slice is:

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

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

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

Using the substitution:

$$u=9-y\,\therefore\,du=-dy$$ we obtain:

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

Using the rule:

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

$$V=\frac{\pi}{4}\int_0^9 u^2\,dy$$

Applying the FTOC, we obtain:

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