0
$\begingroup$

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?

$\endgroup$
2
  • 3
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.