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The region enclosed by the given curves is rotated about the specified line. Find the volume of the resulting solid.

$$x-y=1, y=x^{2}-4 x+3 ; \quad \text { about } y=3$$

$$

21.6 \pi

$$

Applications of Integration

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I want to find the value. And here, looking at my graphs, why equals X minus one? And why equals X squared minus four x plus tree. So to find this falling for responding to know the area, the area of intersection happens here. And then we need to know where we're rotating and this problem were rotating around the horizontal line Y equals three. So as I'm considering this revolution, I hope it's pretty clear that the volume has all of this portion that's missing. And since that parts being taken away, I know this is a washer set up. So for the washer set up, value equals pi times the integral of capital R squared minus lower case R squared. And then we want to use whatever very well is the access. In this case, Y equals three is like using the X axis. Do you have? Those are parallel, So money is DX, which means I need an X one and next to toe label where the area starts and where the area ends. The start of the area happens where the two graphs intersect, and that happens at X equals one, and then the area ends that final limit is where the two graphs meet again. Maybe I can't tell based on my sketch. So a way to find it algebraic Lee is to set your two equations equal to let X minus one equal X squared minus four X plus three. I would say trucked over the X and then add over the one so everything's on one side and then we have to factor this and it factors as X minus one and X minus four, which means the intersections happen at one which we knew from just drawing it and inspecting it and also act for So for my integral, I'm gonna write pie Integral going from 1 to 4. Now, I need to figure out what his capital, R and lower case R capital are is. What if the volume was totally filled in? What if there was nothing missing at all? Well, that would mean all this area would come all the way up and down from the bottom of the area, shaded all the way up to wherever you're rotating the whole way across notice. I feel that all in well, that would be just defined by three minus the parabola upper minus lower there for the area. But be careful that as you subtract the proble to make that lower bound, you use parentheses. So it's subtracting the entire equation. So that's my capital are, and I'm just gonna use parentheses. I don't want to make any mistakes with my signs. So that's my safest bet for writing it correctly. And then for lower case R. I need to decide what was taken away. Well, the stuff taken away is this area between what I've shaded in and where I'm rotating. So the stuff taken away goes from three down to the linear function, which is X minus one. And again, I'm using parentheses there to make sure I'm subtracting the entire second function. So three parentheses X minus one and that whole thing is squared because that's the lower case. Far. So now my inner girl set up is complete. This is a great time. If I have the ability to calculate this integral, I would just calculate it online B In my calculator, however, I could do that instead of having to expand each of those by no meals. So we'll polynomial for the 1st 1 and as I do that I get high still. Outfront times 10 8/5. And depending on the exactness needed. I could also write that as one who ate pie over five or 21.6 over. Sorry, times high. Those were the exact same answer and exact same equivalency.

Oklahoma Baptist University

Applications of Integration