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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.

$$y=1+\sec x, y=3 ; \quad \text { about } y=1$$

$$

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

$$

Applications of Integration

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{'transcript': "we're finding the volume of the solar attained by rotating the region founded by the Y equals one plus six. X one equals three. Rotating around the y equals 19 You can first begin by drying out what y equals X looks like before we actually tackle the way because one post, second X remember that second X is just equal to one over coastline, so we can first draw that out. Should be a pretty familiar curve. The overall shape looks like this, right? And if we be more specific at this point, is one. This is part of the two. This is pie. There's a negative one and this is negative pi over two. So if we take the inverse of that the second X, it's just come a multiplicity of inverse of co sign Children here. Okay, out zero 1/1 is just one. So seeking ex still looks like right here. That's the point. Part of a two second X is undefined. But as you can notice, the cosine X is decreasing. So therefore the second X will increase an approach infinity. So there will be an awesome to hear and likewise on the other side will also increase towards another. Ask them to at pi equal. Uh, y equals negative or X equals negative pi over two. And there will be another shaped like that approaching another. Assam took every single time co sign X cosign X is equal to zero. Seeking X is approaching either positive infinity or an ambulance. Okay. And so that is what? Seeking. Excellent. Now all we have to do is add one. So let's go to a new one, adding one. So instead of one, this is now too. Once again, we are, Yeah, a shape that is approaching towards infinity. And we can call this point right here, however to and it's not really in the actual graph because, as you will later see, um, we don't actually go that far to create the region. Now, if we go back the way it goes, three line is pretty simple. It's just a horizontal line at y equals three. So we can call at this 0.3. That's what we have. So we are rotating in this region about the y equals one line. Mm. Yeah. So this is the way it goes. One line retaining about it. So what? We are doing our creating washers. Now let's find these two points of intersection, and the way we do that is by sending the two equations equal to each other. So we do have one plus C. Connects is equal 23 because we had y equals three and why it goes. One policy connects doing some calculations we get C connects is equal to two. And remember, the second X is just one overcoats annex. So therefore, Cossacks, it's just one half. So X is really just co sign were really our co sign of one half, which means that whichever Number X is whichever, um, value that has a coastline of one half. And if we draw the unit circle, all right, we want to find the points on the unit circle or the angles on the unit circle when the X value is equal to one half, just like right here. And so it's these two points, and it turns out that the co sign of Partner three is one house and the co sign of negative five or three. It's also one half, so these are the two X values, So if you go back here. This is negative power of three, right? This is part of the three. I mean, so those are our two values. And that's why I told you before that the region itself doesn't actually reach the point of X equals five or two because it intersects before then intersects X power three, which is smaller. So now we can come up with the cross sectional area now as a function of X because it changes as exchanges. So when you find the big radius in the small radius, so we're gonna begin by finding the big radius. Now, if you see here the big radius itself, it's pretty constant. The line that's creating the Big Radius is this line right here, which is why it goes three line. So the distance between the like ALS three line and the Y equals one line is always too so bigger. He is always too. The small radius is more complicated because it's this curve that's creating the small radius, and the distance changes at different points of X. So the way we find those small radius is just taking the Y value. So the Wye Valley is just one Post c connects and we are subtracting one because that's the just is there. So we get a distance of just see connects. You can plug that in and pulling up the pie to school is for minus second square. Right? So that is our cross sectional area function. You can go ahead with the interval now. Now, as I showed before, we're integrating from negative pi over three to pi over three and the actual cross sectional area function is pie for numbers so you can score it x the X I want to pull out the pie and evaluate the anti derivatives. The anti driven for is four x the anti derivative of second Squared X, um is 10 because the tangent of the derivative of tangent is just eaten squaring or you're evaluating from maybe a part of the three people, obviously. So when X equals piling three a bit high for power, visiting the tangent of five or three is three. So there was that. Now let's evaluate when it's negative. Pi over three, four times that is millions parliamentary. The tangent of negative pi over three is just negative for three. Yeah, so really, we are adding four power for three. Right there we are subtracting. Think, uh, the Route three. So what we get is that four pi over three plus four by over three is eight pi over three, and we get, um, minus route three minus another three comes minus two. Group three. Yeah, and that is our final answer. Not the way you can simplify it is is if you pull out the two. So you get two pi four pi over three. One is three. Both are acceptable, and this is our final answer."}