Question

Use cylindrical shells to find the volume of the solid obtained by rotating about the x-axis the region under the curve $y = \sqrt{5x}$ from 0 to 5. Solution This problem can be solved using disks. To use shells we relabel the curve $y = \sqrt{5x}$ as $x = $ For rotation about the x-axis we see that a typical shell has radius y, circumference $2\pi y$, and height $V = \int_0^5 (2\pi y)(5 - \frac{y^2}{5}) dy = 2\pi \int_0^5 (5y - \frac{y^3}{5}) dy$ $= 2\pi \left[ \right]_0^5 = $ 

          Use cylindrical shells to find the volume of the solid obtained by rotating about the x-axis the region under the curve $y = \sqrt{5x}$ from 0 to 5.
Solution
This problem can be solved using disks. To use shells we relabel the curve $y = \sqrt{5x}$ as $x = 
$
For rotation about the x-axis we see that a typical shell has radius y, circumference $2\pi y$, and height
$V = \int_0^5 (2\pi y)(5 - \frac{y^2}{5}) dy = 2\pi \int_0^5 (5y - \frac{y^3}{5}) dy$
$= 2\pi \left[ \right]_0^5 = $
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Use cylindrical shells to find the volume of the solid obtained by rotating about the x-axis the region under the curve y = √(5x) from 0 to 5. Solution This problem can be solved using disks. To use shells we relabel the curve y = √(5x) as x = For rotation about the x-axis we see that a typical shell has radius y, circumference 2π y, and height V = ∫0^5 (2π y)(5 - (y^2)/(5)) dy = 2π∫0^5 (5y - (y^3)/(5)) dy = 2π[ ]0^5 =

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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Use cylindrical shells to find the volume of the solid obtained by rotating about the x-axis the region under the curve y=sqrt(5x) from 0 to 5 . Solution This problem can be solved using disks. To use shells we relabel the curve y=sqrt(5x) as x= (i) For rotation about the x-axis we see that a typical shell has radius y, circumference 2pi y, and height . So the volume is V=int_0^5 (2pi y)(5-(1)/(5)y^(2))dy=2pi int_0^5 (5y-(1)/(5)y^(3))dy =2pi []_(0)^(5)= In this problem the disk method is simpler. Need Help? Use cylindrical shells to find the volume of the solid obtained by rotating about the x-axis the region under the curve y = 5x from 0 to 5 Solution This problem can be solved using disks.To use shells we relabel the curve y= 5x as x shell height=5-y2/5 shell radius= X=5 ? For rotation about the x-axis we see that a typical shell has radius y, circumference 2ty, and height .So the volume is In this problem the disk method is simpler. Need Help? R
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Transcript

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00:01 Into the y axis i'm just going to kind of jump into what's going on here is you have this this curve got x equals one and we're going to take all of these regions and we're going to rotate it using shells um and well i hope you can see it's the lateral area of a cylinder which is 2 pi r h so as i'm looking at this problem the first one so number one is going to be 2 pi and then the integral is from 0 to 1.
00:32 Those are my x values.
00:34 The radius of each one of these cylinders is x.
00:38 And the height is that 17x squared dx.
00:42 So if i were doing this problem, i would multiply be 17x cubed.
00:48 So if you take the anti -derivative of that, you have to add one to that exponent, and then you have to divide by your new exponent going from 0 to 1.
00:56 And please don't forget about that 2 pi in front.
00:59 And this is a very time.
01:00 Really nice because if you plug in one right there, one to any power is just one.
01:05 And so i'm going to end up with 17 pi.
01:09 And remember two goes into four twice.
01:12 And that'll be your final answer.
01:13 Now, i am plugging in zero, but it's plugging in zero is just going to give me minus zero.
01:17 So you don't have to write that down.
01:19 So the next one, i don't believe, is much different.
01:22 It's kind of set up the same way.
01:26 Except so because it's such a long problem, i'm just going to jump right into.
01:30 2 pi the bounds they give you are from 2 to 3 the radius is still x and the height of each one of these is 9 over x dx which is actually a really nice problem because these xes will cancel out and i can figure out the integral of 9 it's just 9x and do that from 2 to 3 just make sure you plug in your bounds so 3 times 9 is 27 2 times 9 is 18, which will give me 9, but i have to multiply by that 2 pi.
02:05 So it should be 18 pi...
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