Given $f(x)=x-1$ and $g(x)=x^2-4x+3$. Let $R$ be area of intersection of $f(x)$ and $g(x)$. Find the volume of $R$ rotated around $x$-axis.
First I try to plot the graph of $f(x)$ and $g(x)$ as follows.
Now, I compute the volume of $R$ if rotated around $x$-axis using this formula $$V=\pi \int\limits_{1}^{4} \left(\left(f(x)\right)^2-\left(g(x)\right)^2\right)dx$$ then I obtained $\dfrac{27}{5}\pi$ unit of volume.
Now I confused that volume is correct or wrong. I confused because $R$ intersect the $x$-axis so if $R$ rotated around $x$-axis then the volume is overlap. So my question: Does this formula can correctly compute the volume of this problem? $$V=\pi \int\limits_{1}^{4} \left(\left(f(x)\right)^2-\left(g(x)\right)^2\right)dx$$
The formula should not be used like that. There are parts that do intersect and parts that do not intersect. For example, let's rotate the functions $f(x)=x^2-1$ and $g(x)=-x^2+1$. If you rotate these two around $x$-axis, you will get the same solid, because $f(x)=-g(x)$. It will have a solid of positive volume, but with your formula, you will get that volume is 0. The problem is that parts of your solid intersect, but only some parts. You can try calculating a volume that is bounded by $|g(x)|$, $f(x)$, and $x$-axis (bounded by $|g(x)|$ and $x$-axis on interval $[1,2]$; bounded by $f(x)$ and $|g(x)|$ on interval $[2,4]$). This way, you will not have to worry about the intersection.