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I've recently got the question about rotation around x-axis. You know, where you calculate the volume, when a function is rotated $360^o$ around.

In my books they says that the function should be continuous from $a$ to $b$ and not negative.

This picture show what I mean: continuous non-negative function

And my question here: continuous negative function

I did an example with $f(x)=-x^2+5x+10$ and $g(x)=2x$ where I have to find the volume of $f$ and $g$ where both functions are rotated around $x-axis$.

My example calculations: $V=\pi\int_{-2}^{5}(f(x)^2-g(x)^2)dx=\frac{27097}{30}\pi$ and a graph (not 3D) here: graph of the functions

I hope your guys understand me and can help me to answer the question.

Amir Hassan
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1 Answers1

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I assume you are finding the volume bounded by $R=f(x)$ on the outside and $R=g(x)$ on the inside.

If the function $f(x)$ is positive and $g(x)$ is negative at that same value of $x$, then the disk-shaped volume-element "bounded" by the rotations of $f$ and $g$ will have a radius of the larger of $f(x)$ and $g(x)$. But the integral you get by just applying the formula would give the same thing as the disk-with-a-hole-shaped volume-element bounded by the two positive radii $f(x)$ and $-g(x)$. Obviously, these are not equivalent.

For just a single function rotated around the $x$ axis, you actually can work with positive and negative $f(x)$ in the same volume integral without encountering this error.

Mark Fischler
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