While practicing finding the volume of shapes bounded by a function and the $x$-axis by rotating the shape around the $y$-axis and taking an integral (the "washer" or "pipe" method) I became confused as to why the shape used to calculate volume "stops" at the $x$ axis, and isn't a part of the calculation.

To explain further, if the shape is being discretized in the $x$ direction ($dx$), then the integral will be set up to be calculated between two $x$ values, call them $a$ and $b$. This restricts the calculation to be determining the area only between the $x$ values of $a$ and $b$. In the $y$ direction, however, there is no such restriction, and so far I have been assuming that the $y$ component of the calculation "begin" at the function being integrated and "end" at the $x$-axis. I do not understand, however, the mathematics behind why I get the correct answers for the volume of the rotated shape if I am ignoring the area that extends from the $x$ axis to negative infinity in the $y$ direction.

A simple example that can help illustrate my question is finding the volume of the shape defined by $y=x^2$ between $0< x < 2$ rotated around the $y$ axis. Why, in this example, is the area below the $x$ axis not considered?

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    $\begingroup$ There isn't any area under the $x$ axis in the case $y=x^2$. Did you intend to write another function? $\endgroup$
    – Element118
    Dec 22 '15 at 3:11
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    $\begingroup$ The problem should specify the region more carefully, as in "the region below $y=x^2$, and above the $x$-axis, from $x=0$ to $x=2$, is rotated $\dots$." $\endgroup$ Dec 22 '15 at 4:04

In your example there is no area below the $x$ axis, but if your function were $y=x-1$ from $x=0$ to $x=2$, rotated around the $y$ axis, you are considering a cone and definitely want the area below the $x$ axis


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