Surface Area of Solid of Revolution (about x-axis, formula explained)
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- čas přidán 29. 10. 2018
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Surface Area of Solid of Revolution,
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blackpenredpen | 曹老師
∫2π(y)dL = surface area of a solid by revolution
∫2π(we)dL = surface area of a solid by Bolshevik revolution
Kostas T. Expain?
i hate that i laughed at this
Ujwal 9000 can you explain what's going on?
@@seeme24x7 Bolsheviks were a communist group in russia during the early 20th Century. In communism,their belief is that property and money should be shared equally by all. So they preached the idea of 'we' and 'us' instead of 'me' and 'i' .So notice how in that integral he substitutes y with we and calls it a bolshevik revolution(a political revolution or a revolution as in rotation).Thats the joke.
When I dont understand the lecture notes, I watch blackpenredpen
I'm from Egypt, and I really love your way of solving problems. Thank you very much ❤👷♂️
My Calc teacher loves your videos! Great work
Thank you so much for this amazing explanation.
Very nice video
Loved it,😀
Omg,u solve my assignment for calculus ,thank younso much
Your "s" and your integral signs look similar - I just learned this month that the integral sign came from the style of the S in those years - because an integral is a Sum.
Carry on. You are very helpful!
Your videos are wonderful, I like your all video because i am a highschool student but still i can understand because of your great explanations .
really appreciate it thanks
Tremendous bro, love your style at teaching, keep it going!!
By the way bro I'm still at school, at the eleventh class 😂and didn't take anything you mentioned in the video, at my school, but I'm still amazed!
Reason for why dl for surface area and dx for volume:
For simplicity, imagine a straight line passing through origin as your curve. Now rotate it around X axis to create surface of revolution. Now, cut the surface at y=0 and z=-f(x). Open the surface without disturbing z=f(x). Now, shift your view to the plane of z=f(x) and y axis. Now, the problem is so simple. Area under your curve is integral 2pif(x)ds.
If you project your cut surface onto XY plane, then the surface area in XY plane is integral 2pif(x)dx.
=integral 2pif(x)ds*(dx/ds)
=integral 2pif(x)ds * (cos theta) -------(1)
Theta is the angle made by your line with the X axis.
Eqn(1) makes sense because we are just projecting our surface onto the XY plane.😊
Now we can replace the line with any curve.
Imagine the same situation as above. Now the area in YZ plane is pif(x)^2. Now, we can make it a 2 dimensional problem.
Create a coordinate system in which my Y axis has the area in YZ(previous) plane and X axis has same length as in previous X axis. Now, the area under the curve should give the volume of the surface.
Therefore, volume = integral pif(x)^2dx
Remeber, we don't take any slant height in area problem. Its just integral f(x)dx. That is why the 'dx' here is not the slant height.
THANK YOU SO MUCH
Thankyou for video. It cleared my doubt of why we are not taking dx element here. I was struggling with that.
Thank you so much
Awesome explanation, gotta say
I always liked you for your videos, but now I like you for that shirt too!
Very nice
Thank you. The video clearly explains why the dL will be a slanted line and not a horizontal line, in addition to providing a simple explanation of the formula. Very straightforward and useful video. Liked!
did u actually get that? if u dif can u explain why cuz i don't get it yet
@@BilalAhmed-on4kd I guess I got it 4 years ago when I was taking calc. Now I forgot all this shit 😂
you are great bro
Hello,
How to derive the area functional for a surface of revolution around the x-axis from the general area functional for the special case of a surface of revolution whose axis coinciding with the x-axis?
Good stuff. Most sites try to avoid the dy world because it is tricky, not as straightforward as the dx world. You did this which is great! but unfortunately at the end you did not elaborate what the dx/dy is, the key part by solving with dy.
I didn't quite get the dL stuff at the and, how did the length of the curve (the square root of...) cropped up?
thanks for an easy and short way of deriving concepts love from Hindustan ❤❤❤
thanks
nice
i hope you're having a great day
I have a question: Why can't it be integrated with respect to dx? dx is a small change and we would still be adding all possible circumferences that change with respect to the function.
i am also having the same doubt but doing so is leading to wrong answers
@@tejaskataredx is a small change horizontally while dl accounts for the small change in the distance between two points so it takes care of both the horizontal and vertical change. There is another video by bprp explaining arc length where he explains why we use dl
make a video on the y-axis!
I always wonder why we use dl instead of using dx like solving the volumn of the revolution?
same, i am confused, pls help.
@@user-lc6jq1hi1r It is jsut why I came to see the video
For simplicity, imagine a straight line passing through origin as your curve. Now rotate it around X axis to create surface of revolution. Now, cut the surface at y=0 and z=-f(x). Open the surface without disturbing z=f(x). Now, shift your view to the plane of z=f(x) and y axis. Now, the problem is so simple. Area under your curve is integral 2pif(x)ds.
If you project your cut surface onto XY plane, then the surface area in XY plane is integral 2pif(x)dx.
=integral 2pif(x)ds*(dx/ds)
=integral 2pif(x)ds * (cos theta) -------(1)
Theta is the angle made by your line with the X axis.
Eqn(1) makes sense because we are just projecting our surface onto the XY plane.😊
Now we can replace the line with any curve.
Imagine the same situation as above. Now the area in YZ plane is pif(x)^2. Now, we can make it a 2 dimensional problem.
Create a coordinate system in which my Y axis has the area in YZ(previous) plane and X axis has same length as in previous X axis. Now, the area under the curve should give the volume of the surface.
Therefore, volume = integral pif(x)^2dx
Remeber, we don't take any slant height in area problem. Its just integral f(x)dx. That is why the 'dx' here is not the slant height.
Hey I’m not sure if you have physics knowledge but I really need help with buoyancy of shapes with parabolic surfaces
You're the MVP 👊
Do you have any of these Surface area videos but ones that revolve around y-axis with give x equation. ones that rotate about x axis with given y-equations etc... Also looking for arc length if you have like a playlist. Thank you. Exam on Wednesday!
What if the solid is rotating around a non-axis point like y=1?
Hi, i love your videos 😍
Would You please talk about riemann's method for trigonometric functions?
From what I know, that can only be done with Taylor´s expansion.
I tried the same thing some months ago, comparing definite integral using just direct Riemann´s definition and using Riemann´s method but with the Taylor expansion on a Python program, and it´s pretty cool.
Why do you use the arc length for surface area but not for volume
For simplicity, imagine a straight line passing through origin as your curve. Now rotate it around X axis to create surface of revolution. Now, cut the surface at y=0 and z=-f(x). Open the surface without disturbing z=f(x). Now, shift your view to the plane of z=f(x) and y axis. Now, the problem is so simple. Area under your curve is integral 2pif(x)ds.
If you project your cut surface onto XY plane, then the surface area in XY plane is integral 2pif(x)dx.
=integral 2pif(x)ds*(dx/ds)
=integral 2pif(x)ds * (cos theta) -------(1)
Theta is the angle made by your line with the X axis.
Eqn(1) makes sense because we are just projecting our surface onto the XY plane.😊
Now we can replace the line with any curve.
Imagine the same situation as above. Now the area in YZ plane is pif(x)^2. Now, we can make it a 2 dimensional problem.
Create a coordinate system in which my Y axis has the area in YZ(previous) plane and X axis has same length as in previous X axis. Now, the area under the curve should give the volume of the surface.
Therefore, volume = integral pif(x)^2dx
Remeber, we don't take any slant height in area problem. Its just integral f(x)dx. That is why the 'dx' here is not the slant height.
Find the surface area due to the rotation of the area between f(x)=x^3 and g(x)=x
Ok I've watched all the videos in the Integral formulas playlist and I still can't understand why a dx integral is enough for the volume but a dl integral is required for the area.
For that matter I don't understand why is it just a cylinder when you calculate the volume but must be a band of a cone when you calculate it's surface area! Please help!
Basically, you're taking a section of a given length and sweeping it around, which requires knowing the arc length, which is dependent on the slope at every point on the curve. Think of it as the graph "reaching" some distance along the x-axis. A line with a steep slope has to be longer to reach as far as a line with a very shallow slope. The same concept applies to the arc length, except now you have to account for all the changes in slope. Thus, you can use triangular approximation to find the slopes all throughout the curve. Let's take a cylinder as an example. Place it on the x-axis like a roll of toilet paper on its holder. If you exclude the circular faces of a cylinder, the sides are straight, so the slope is zero, meaning the arc length is just h because there's no bumps or kinks adding extra length, hence how the surface area of an unrolled cylinder is 2(pi)(r)(h). If the cylinder had a bulge in the middle, it's surface area would be greater than that, and to do that, you'd need a radius that changes (that's your function), and the description of how your slope changes to fit that changing radius (dL).
Cylinder approximation is for volume because it's easy and intuitive, assuming you get what a definite integral is trying to express. Remember, it's a Riemann Sum, which adds up infinitely many shapes that all pack together to give you what you're looking for, be it the volume of an object (cylinders and washers can pack together nicely and fill in all the space in a shape, so having enough of them and adding up their volumes gives you the volume of this complicated shape), the surface area of an object (imagine putting spaghetti on a vase so they all line up from top to bottom with the vase, then take them off and measure the total area they cover- that's basically what we're doing here, just with infinitely many strands of "spaghetti"), or various values in physics.
For simplicity, imagine a straight line passing through origin as your curve. Now rotate it around X axis to create surface of revolution. Now, cut the surface at y=0 and z=-f(x). Open the surface without disturbing z=f(x). Now, shift your view to the plane of z=f(x) and y axis. Now, the problem is so simple. Area under your curve is integral 2pif(x)ds.
If you project your cut surface onto XY plane, then the surface area in XY plane is integral 2pif(x)dx.
=integral 2pif(x)ds*(dx/ds)
=integral 2pif(x)ds * (cos theta) -------(1)
Theta is the angle made by your line with the X axis.
Eqn(1) makes sense because we are just projecting our surface onto the XY plane.😊
Now we can replace the line with any curve.
Imagine the same situation as above. Now the area in YZ plane is pif(x)^2. Now, we can make it a 2 dimensional problem.
Create a coordinate system in which my Y axis has the area in YZ(previous) plane and X axis has same length as in previous X axis. Now, the area under the curve should give the volume of the surface.
Therefore, volume = integral pif(x)^2dx
Remeber, we don't take any slant height in area problem. Its just integral f(x)dx. That is why the 'dx' here is not the slant height.
First to say , I'm a highschool student and don't know much higher mathematics
Can I ask why the circumference of disc ( 2•pi•y•dl ) is considered as circumference
Of the circle instead of surface area of a frustum - the two bases ?
Is it because "dl" is too small to consider a height ?
Is the surface area on the outside or the inside of the object?
Why dl got replaced by (1+dy/dx)sqrd2
You make it look so easy😏😏. How do you get so good?
Why did you equalize dl to root 1+ derivative^2?
ok
For y=f(x) you get integral 2*Pi*x*sqrt(1+y'^2)dx and for x=g(y) integral 2*Pi*g(y)*sqrt(1+x'^2)dy. It's again funnier when both x and y are functions in terms of time.
Someone who recommends me a book about precalculus I would like to know limits or integers to study other books for example theory of numbers I have been searching for it but I can't find a good book.
If I wrote bad a word, sorry, I am learning english xd and good video blackpenredpen you explain so good, greetings and still doing it :)
Cool
But why is a straight line enough in the case of the volume, then?
C H E C K D I S O U T!
Just noticed that hes wearing a WWF(/E) shirt
Yup! : )
It is not clear for me why we should approximate surface area with a piece of a cone and the volume with a cylinder. Shouldn't both use the same idea?
For the surface area, we approximate part of the arc length with a straight line, which when rotated, becomes a frustum (which is part of a cone). And the idea is that dx and dy are small enough so that this approximation is accurate (hence, why we integrate and not sum).
When you rotate a continuous function around an axis to find the volume, you're considering each disk slice and then summing them up. Each disk slice is cylindrical.
Hopefully, that clears some things up :)
oh, now I got it.
I did it in a slightly different way. First, I imagined some ring with radius y somewhere on he solid. Then, the side of such ring is 2pi y. Then I need to add all those rings with negligible thickness. However, I can't do that simply in terms of x. I can imagine the solid is made of fabric, and I can stretch it to get a frustum. Then I integrate through the entire arc length, which is the dl and everything fits.
Good point. Initially I was thinking in the same way. Because we are dealing with infinite number of cylinders. However, indeed dl is crucial for surface integrals. From the other hand we would obtain infinitely small error for volume integration.
For simplicity, imagine a straight line passing through origin as your curve. Now rotate it around X axis to create surface of revolution. Now, cut the surface at y=0 and z=-f(x). Open the surface without disturbing z=f(x). Now, shift your view to the plane of z=f(x) and y axis. Now, the problem is so simple. Area under your curve is integral 2pif(x)ds.
If you project your cut surface onto XY plane, then the surface area in XY plane is integral 2pif(x)dx.
=integral 2pif(x)ds*(dx/ds)
=integral 2pif(x)ds * (cos theta) -------(1)
Theta is the angle made by your line with the X axis.
Eqn(1) makes sense because we are just projecting our surface onto the XY plane.😊
Now we can replace the line with any curve.
Imagine the same situation as above. Now the area in YZ plane is pif(x)^2. Now, we can make it a 2 dimensional problem.
Create a coordinate system in which my Y axis has the area in YZ(previous) plane and X axis has same length as in previous X axis. Now, the area under the curve should give the volume of the surface.
Therefore, volume = integral pif(x)^2dx
Remeber, we don't take any slant height in area problem. Its just integral f(x)dx. That is why the 'dx' here is not the slant height.
Hi, I hope I am not too late to write this,
Why is y=f(x)?
I thought the radius of revolution would be the average of f(x) and f(x+dx), does this show a different result? Is this incorrect? What's the difference?
somebody give me a hell yeah!!!
Excellent!
BTW - Is your choice of Demonstration Props a sad reflection of your funding or merely a chance to include some 'toilet humour' for a bit of fun? Lol
It's wonderful what you are doing!
Find surface area by rotate y=erf(x) about x axis
x from -1 to 1
Can you also teach on how to get the volume of revolution of y=coshx? : )
2:37 Why it’s 2pi y ?
Yeah
hey,is it possible to integrate sqtr(x^3 -1)?
Not with elementary functions, but it is related to the three Elliptic Integral functions, or Hypergeometric functions.
That's Stone Cold Steve Austin T-shirt. Are you a WWE fan too??
Pranesh Pyara Shrestha
Finally someone recognized the shirt!!
Pranesh Pyara Shrestha
I watched it from time to time. 10 years ago was better tho in my opinion.
@@blackpenredpen I am a huge WWE fan. I have that t-shirt too
@@blackpenredpen WWE World Cup is going to be held this Friday, You know?
@@blackpenredpen hahahahahh I wasn't sure if it was SCSA at first but Walla it was. Good taste.
❤❤❤❤
Excuse me,I know it's dumb to ask but,why don't we get the correct answer when we use dx instead of dL,I mean, they're both approaching to 0 so ...there aren't many different between dx and dL right?
Is it something like the Jacobian matrix?
Can someone explain to me why we don’t just use dx instead of arc length? As all of you know the surface area becomes more accurate as dx reaches 0 so it seems to me that using dx will give the same result.
suppose you have a line of gradient 5000, the sum of the dx values clearly is not the true value of arc length
For simplicity, imagine a straight line passing through origin as your curve. Now rotate it around X axis to create surface of revolution. Now, cut the surface at y=0 and z=-f(x). Open the surface without disturbing z=f(x). Now, shift your view to the plane of z=f(x) and y axis. Now, the problem is so simple. Area under your curve is integral 2pif(x)ds.
If you project your cut surface onto XY plane, then the surface area in XY plane is integral 2pif(x)dx.
=integral 2pif(x)ds*(dx/ds)
=integral 2pif(x)ds * (cos theta) -------(1)
Theta is the angle made by your line with the X axis.
Eqn(1) makes sense because we are just projecting our surface onto the XY plane.😊
Now we can replace the line with any curve.
Imagine the same situation as above. Now the area in YZ plane is pif(x)^2. Now, we can make it a 2 dimensional problem.
Create a coordinate system in which my Y axis has the area in YZ(previous) plane and X axis has same length as in previous X axis. Now, the area under the curve should give the volume of the surface.
Therefore, volume = integral pif(x)^2dx
Remeber, we don't take any slant height in area problem. Its just integral f(x)dx. That is why the 'dx' here is not the slant height.
You used dl for surface area but dx for volume 😕
You seem to be using two different methods for the same question why could you not do the sum of the rings (circumference with the thickness dx) as you did the sum of the discs for volume?
For simplicity, imagine a straight line passing through origin as your curve. Now rotate it around X axis to create surface of revolution. Now, cut the surface at y=0 and z=-f(x). Open the surface without disturbing z=f(x). Now, shift your view to the plane of z=f(x) and y axis. Now, the problem is so simple. Area under your curve is integral 2pif(x)ds.
If you project your cut surface onto XY plane, then the surface area in XY plane is integral 2pif(x)dx.
=integral 2pif(x)ds*(dx/ds)
=integral 2pif(x)ds * (cos theta) -------(1)
Theta is the angle made by your line with the X axis.
Eqn(1) makes sense because we are just projecting our surface onto the XY plane.😊
Now we can replace the line with any curve.
Imagine the same situation as above. Now the area in YZ plane is pif(x)^2. Now, we can make it a 2 dimensional problem.
Create a coordinate system in which my Y axis has the area in YZ(previous) plane and X axis has same length as in previous X axis. Now, the area under the curve should give the volume of the surface.
Therefore, volume = integral pif(x)^2dx
Remeber, we don't take any slant height in area problem. Its just integral f(x)dx. That is why the 'dx' here is not the slant height.
Why can't the formula for the surface area of a cylinder be used?
S(cylinder) = 2πRL
R = Radius = y = f(x)
L = Length of the function = Arc Length = {1+[f'(x)]^2}^(1/2) * Δx
Surface area of the function is the limit (as n approaches infinity) of the summation (from i=1 to i=n) of {2π*f(x)*[1+f'(x)^2]^(1/2)*Δx}
S=2π∫f(x)*[1+f'(x)^2]^(1/2) dx
Can anybody explain how dl=root(1+(dy/dx)^2) happens?
dl = sqrt((dx)^2+(dy)^2) = sqrt((dx^2)(1+(dy/dx)^2)= sqrt(1+(dy/dx)^2).dx
@@joluju2375 Thank you.
it is the formula for arc length which is in this video the red line on function
@@punaydang2948 Thank you.
You could just do a surface integral
Can we not use dx instead of dL??????
i think he writes dL because dL is the small value increment ON the curve, whereas dx is a small increment on the x-axis.
I am not sure though
for anyone curious, it goes back to his video on deriving arc length in a non-rigorous/more intuitive way. here's the video: czcams.com/video/PK7HZiFG_VI/video.html
in short, it's really an application of the pythagorean theorem. dL is a differential that represents an infinitesimally small portion of the actual arc length (the "hypotenuse"), dy is the infinitesimally small change in y of this right triangle, and dx is the infinitesimally small change in x of the right triangle. dx would just represent the lower base of that right triangle, so it wouldn't represent the arc length.
@@tb2748 Why don't we use a dL when using the disk method of integration? It seems like using a dx here wouldn't matter because the surface area is an approximation which gets better as dx approaches zero. In short, why don't we involve the arclength when finding volume using the disk method?
For simplicity, imagine a straight line passing through origin as your curve. Now rotate it around X axis to create surface of revolution. Now, cut the surface at y=0 and z=-f(x). Open the surface without disturbing z=f(x). Now, shift your view to the plane of z=f(x) and y axis. Now, the problem is so simple. Area under your curve is integral 2pif(x)ds.
If you project your cut surface onto XY plane, then the surface area in XY plane is integral 2pif(x)dx.
=integral 2pif(x)ds*(dx/ds)
=integral 2pif(x)ds * (cos theta) -------(1)
Theta is the angle made by your line with the X axis.
Eqn(1) makes sense because we are just projecting our surface onto the XY plane.😊
Now we can replace the line with any curve.
Imagine the same situation as above. Now the area in YZ plane is pif(x)^2. Now, we can make it a 2 dimensional problem.
Create a coordinate system in which my Y axis has the area in YZ(previous) plane and X axis has same length as in previous X axis. Now, the area under the curve should give the volume of the surface.
Therefore, volume = integral pif(x)^2dx
Remeber, we don't take any slant height in area problem. Its just integral f(x)dx. That is why the 'dx' here is not the slant height.
The last part is very confusing, 2piy should have been replaced by 2pix
I suppose you could say the circumference is 2 ply y I hate myself
When's the channel's name gonna change to blackpenredpenbluepen? :D
What about the surface area of the disks of your a and b boundary conditions? Your stated problem was "what is the surface area of a solid of revolution?" You did not finish answering your own question.
This is a little late but I had the same question and found that we're only taking the lateral area of the revolved curve. In other words, we're just not considering the area of the two disks.
We call "the part of a cone" a frustum.
Hi, why can't we consider dL to be equal to dx, like we did in the volume part we can make dL so small that it is equal to dx? I know this video touched on this concept a bit but I am still confused
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