Volume of a solid?

$$\begin{array}{rrrrrrrr} x & 0 & 0.5 & 1.0 & 1.5 & 2.0 & 2.5 & 3.0 \\ f(x) & 2 & 1.3 & 0.9 & 0.6 & 0.7 & 1.1 & 1.9 \end{array}$$

Find a formula for the volume $V$ of the solid whose base is the region bounded by $y = f(x)$, the $x$-axis, and the line $x = 3$ and its cross-sections perpendicular to the $x$-axis are semicircles.**

So, I plotted the points and got a graph that looks something like this:

Now to start on actually solving the problem.

So I figure that we should break the region up into a small $dx$ pieces, and just sum up all of these pieces using an integral.

However, I'm having trouble figuring our what the area of each piece will be. Any help?

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math.lsa.umich.edu/courses/116/TeamHw/TeamHw3/TeamHw3.pdf Please don't answer questions like this... – Brooke Ullery Feb 3 '12 at 2:44

If I am understanding the geometry right, the semicircles have radius $f(x)/2$, and therefore area $(\pi/8)(f(x))^2$. Call this area $g(x)$.

We want $\int_0^3 g(x)\,dx$. But since we do not have a formula for $f(x)$, we do not have one for $g(x)$.

You are now very likely intended to use a numerical integration method.

From the given values of $f(x)$, you can compute $g(x)$ at $0$, $0.5$, and so on up to $3$. So you have a subdivision of the interval of integration into $6$ equal parts of width $\Delta x$, where $\Delta x=0.5$. Now it is straightforward to use the Trapezoidal Rule, or Simpson's Rule. If numerical integration has not been discussed yet, you can use a Riemann sum to approximate the integral. The error is then likely to be larger than if you use Simpson's Rule, or the Trapezoidal Rule.

Remark: To cut down on the wear and tear on your calculator, and on the likelihood of keying errors, you should alter the computational procedure slightly. Let $h(x)=(f(x))^2$. Then we want $\frac{\pi}{8}\int_0^3 h(x)\,dx$. Now we only need to find the $(f(x_i))^2$, process using say Simpson's rule, and multiply by $\pi/8$ to get the final answer. Squaring of $2$, $1.3$, and so on up to $1.9$ can even be done in one's head.

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Thank you, extremely helpful reply. I was just wondering how you got the f(x)/2 to be the radius? – Jason Jan 27 '12 at 0:39
The $3$-D shape has a flat base in the $x$-$y$ plane. Look at a slice perpendicular to the $x$-axis. It is a semicircle. The "flat" part of the semicircle is a straight line that extends from $(x,0)$ to $(x,f(x))$. So the diameter of the semicircle is $f(x)$ ("$y$"). To find the radius, divide by $2$. I would like to do better, by drawing a good picture of the picture in my head. I hope that you will be able to see a similar picture! – André Nicolas Jan 27 '12 at 0:48