Blaze Jensen

2023-03-03

Find the volumes of the solids obtained by rotating the region bounded by the curves $y={x}^{2}$ and $y=x$ about the following lines. The x-axis.

Hunter Hendricks

Finding the x values where our two functions intersect must come first. Setting the two functions equal to one another and finding x will accomplish this:
${x}^{2}=x$
${x}^{2}-x=0$
$x\left(x-1\right)=0$
$x=0,1$
These x values mean the region bounded by functions $y={x}^{2}$ and $y=x$ occurs between x = 0 and x = 1.
We will apply the formula: to find the volume about the x axis. $V={\int }_{a}^{b}\pi \left\{\left[{f\left(x\right)}^{2}\right]-\left[{g\left(x\right)}^{2}\right]\right\}dx$
Our two functions are represented by f(x) and g(x), with f(x) being the larger function.
Simply choose a number between 0 and 1, and plug it into both of your functions to see which of them is larger. Your larger function is the one that produces the larger number.
If we plug, say $\frac{1}{2}$ into our two functions for example, we will get:
$y=x=\frac{1}{2}$
$y={x}^{2}=\frac{1}{4}$
Thus, our larger function is y=x.
Our integral should look like this:
${\int }_{0}^{1}\pi \left[{\left(x\right)}^{2}-{\left({x}^{2}\right)}^{2}\right]dx$
Remember : since the region bound by our two curves occurred between x = 0 and x=1, then 0 and 1 are our lower and upper bounds, respectively.
Since pi is a constant, we can bring it out: $\pi {\int }_{0}^{1}\left[\left({x}^{2}\right)-{\left({x}^{2}\right)}^{2}\right]dx$
Solving this simple integral will give us: $\pi {\left[\frac{{x}^{3}}{3}-\frac{{x}^{5}}{5}\right]}_{0}^{1}$
=$\pi \left[\left(\frac{{1}^{3}}{3}-\frac{{1}^{5}}{5}\right)-0\right]$
=$\frac{2}{15}\pi$

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