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Find the volume of the solid obtained by rotating…

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Problem 1 Problem 2 Problem 3 Problem 4 Problem 5 Problem 6 Problem 7 Problem 8 Problem 9 Problem 10 Problem 11 Problem 12 Problem 13 Problem 14 Problem 15 Problem 16 Problem 17 Problem 18 Problem 19 Problem 20 Problem 21 Problem 22 Problem 23 Problem 24 Problem 25 Problem 26 Problem 27 Problem 28 Problem 29 Problem 30 Problem 31 Problem 32 Problem 33 Problem 34 Problem 35 Problem 36 Problem 37 Problem 38 Problem 39 Problem 40 Problem 41 Problem 42 Problem 43 Problem 44 Problem 45 Problem 46 Problem 47 Problem 48 Problem 49 Problem 50 Problem 51 Problem 52 Problem 53 Problem 54

Problem 1 Medium Difficulty

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.
$$y=2-\frac{1}{2} x, y=0, x=1, x=2 ; \text { about the } x-axis$$

Answer

19 $\mathrm{pi} / 12$

Related Courses

Calculus 2 / BC

Essential Calculus Early Transcendentals

Chapter 7

APPLICATIONS OF INTEGRATION

Section 2

Volumes

Related Topics

Applications of Integration

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April 12, 2021

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Watch More Solved Questions in Chapter 7

Problem 1
Problem 2
Problem 3
Problem 4
Problem 5
Problem 6
Problem 7
Problem 8
Problem 9
Problem 10
Problem 11
Problem 12
Problem 13
Problem 14
Problem 15
Problem 16
Problem 17
Problem 18
Problem 19
Problem 20
Problem 21
Problem 22
Problem 23
Problem 24
Problem 25
Problem 26
Problem 27
Problem 28
Problem 29
Problem 30
Problem 31
Problem 32
Problem 33
Problem 34
Problem 35
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Problem 37
Problem 38
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Problem 40
Problem 41
Problem 42
Problem 43
Problem 44
Problem 45
Problem 46
Problem 47
Problem 48
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Problem 53
Problem 54

Video Transcript

do you find the volume of the solid that is rotated of the region, bounded by why equals two minus 1/2 X y equals zero X equals one X equals two and we're going around the x axis. So first of all, let's consider that linear function to minus 1/2 X. This has I intercept of two and then a slope of negative 1/2 so down 1/2 down, 1/2 would make this line. It's about a straight. I could draw it and then we're also looking at y equals zero. So that's gonna be your X axis. We're looking at X is wine, and we're looking at X is too. So the region that's bounded by that linear function X is one excess to And why zeros this part of shaded in green Now for rotating it around the X axis, Then I always want to ask myself, Is there any volume there any space there that needs to be cut out where that's missing? And the answer here is know, as we rotate this, it's gonna be a totally solid figure. I know this because every part of that shaded region and green is coming right up against the exact access. So because it's that solid figure, I know for sure it's a disk method. So for the disk method for volume, we have high R squared and we want to look for the ex boundaries. We want to have it in terms of X with our in a girl because we're going around the X axis. Thankfully, we're given the limits for X. The smallest X value is one the greatest Xlu is, too. That's what's bounding this volume. So I'm gonna put one into on my integral and then to fill in our you're thinking of the Radius. So if I'm looking at my figure, the radius is how far I travel from the center. So where I'm spinning Teoh, you know that that edge of the shape and that are here is determined by going from the linear function to minus 1/2 X down to the axis. So I always like to think of if I'm in terms of X is I like to think of it as upper minus lower. The lower function here, though, is just zero. Because it's the Axis, it's helpful to think upper minus lower for if I have shifted this somewhere else into the plane, so we're gonna fill into minus 1/2 X as that are term here. So then we want to calculate this in a rural. It's pretty easy to go through by hand. If you would just expand that by no meal. So if you just multiply and then you can just do each term by term, so multiplying that out and then going term by term Or if you have a calculation program, you can find this integral total. The integral here is 19/12 so I wouldn't leave my answer exact and call it 19 pi over 12.

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