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The region enclosed between the curves $y=x^2−2x+3$ and $y=x+1$ is now rotated about the $x$-axis to form a solid of revolution. Find its volume.

I have tried integration but i am unsure what area the volume would be bounded by, is it $a=1$ and $b=2$ for the integral or do i need to do two separate integrals for the different graphs. My integral so far is

$\displaystyle V= \int_a^b \pi(x+1)^2 - \pi(x^2-2x+3)^2 \, dx$ - Is this correct for the integral?

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You can set the two functions equal to each other to find points of intersection:

$$x^2-2x+3=x+1$$

And solve for $x$; you will get $x=1$ and $x=2$. So now you can say $a=1$ and $b=2$ to be your limits of integration. (What you did in your post is fine.)

As far as splitting the integrals you suggested, that might help, in terms of organization of work. (Note that I'm using $V_1$ and $V_2$ arbitrarily; you can call these anything else you want.) $$V_1=\int_1^2 \pi(x+1)^2 \, dx$$ and $$V_2=\int_1^2 \pi(x^2-2x+3)^2 \, dx$$ and $V=V_1-V_2$.

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  • $\begingroup$ So they don't actually subtract from each either but the integrals are added? and would V2 have the same a and b? $\endgroup$ – ojando Oct 7 '14 at 5:51
  • $\begingroup$ No, you were right; they were subtracted. I made a mistake (sorry!). But yes, $V_1$ and $V_2$ (I simply called them those) have the same $a$ and $b$, due to the linearity of integration. $\endgroup$ – Cookie Oct 7 '14 at 5:52
  • $\begingroup$ can the π be left out the front when integrating? $\endgroup$ – ojando Oct 7 '14 at 6:16
  • $\begingroup$ @JessicaErin no. $\endgroup$ – rae306 Oct 7 '14 at 6:25

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