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# Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. Sketch the region, the solid, and a typical disk or washer.$y = 1 + \sec x$ , $y = 3$ ; about $y = 1$

## 2$\pi\left(\frac{4 \pi}{3}-\sqrt{3}\right)$

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Applications of Integration

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were given curves and a line and rest to find the volume of the solid obtained by rotating the region bounded by these curves. About this line. We have to sketch the region the solid and the typical disk or washer. The curves are y equals one plus the second of x. And Y equals three. Like. And the line we're rotating is about why equals one. Well, first I'm going to sketch the region to do this. You should know how to sketch Y equals seek an fx. This is a standard function and why equals one plus. Speaking of exes just a transformation of the graph up one unit. So mm That was last year we start off with the Y intercept. Well no, you got a bunch of free swag. You got like a Subaru backpack again at two and then we go like this. It really is nice. When is the first order we have a vertical ascent tote. However, Which I'll draw in green here. Mhm. Yeah. Yeah. Yeah. Mhm. These vertical assam totes. Thank you. He was. It's Mhm. These are a plus or -9/2 I mean yeah. Just it's not a good impression. This is what's that And the horizontal line is why equals three. Which is this line. And so this region in red, this is the region we're going to rotate to form are solid. So the question is, what are these points here where a line intersects the curve? I'm often being. Mhm. Well this is the solution to the equation. Sorry, there are enough One plus the seeking of x equals three. So the seeking of X Equals two. In other words the co sign of x equals one half. So that in this case we're really just looking at two values of X. X equals I A negative pi over three and positive pi over three. So these points have coordinates. Hi over three. Three. And the negative pi over three. Three. Yes I love hearing. In fact I'm gonna close my eyes right? Yes Okay now I'll draw the region itself formed from rotating. Well we rotate about the line y equals one which is this green line. So when we rotate reflect this region first it looks something like this. I said in blue. That's so I know that's one of the things I'm going to ask him how much headed. Right I don't have to be on the show adam. Can you chill out man? I'll just stop raising your mood. Let me be. Well yes. Yeah. Now looking at the solid it's clear that we have washers look like this. Do that tell you because if we if we don't make that parking then it's then we were yeah there's nothing else if I could be that we will presario and it's clear that the washer has an inner radius and outer radius Go ahead 1/2 Tom Yeah you'll never produce anything. The inner radius is one Plus II of X -1. That we would need to do anything happen that would sort of put a nice little bow which is just the second of X. And the outer radius. March here fires. This is equal to three -1 which is to So using the watch your method we have that are volume the is equal to pi times the integral from we see X goes from negative pi over 32 positive pi over three. I'm not a hater. I care just wanted some seat And then we have the outer radius two squared minus the inner radius second of x squared dx. This is a little harder to integrate so I'll walk you through it. This is equal to first of all, seeking is even function. Yeah, so this whole function is even and this is the same as two pi times the integral from zero to pi over three of four minus seeking squared of x. Dx I am a friend of. Now if we integrate this is the same as two pi times four X whenever minus the tangent of X from zero to pi over three. This is equal to if you evaluate honestly two pi times four pi over three minus route three, this is our volume.

Ohio State University

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Applications of Integration

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