❖ Double Integral Using Polar Coordinates - Part 1 of 3 ❖
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- Опубликовано: 5 фев 2025
- Master Double Integrals Using Polar Coordinates: Setting Up for Volume Calculations Over Circular Regions!
In this video, we'll explore how to use polar coordinates to set up a double integral for finding the volume underneath a plane and above a circular region. Our limits of integration come from our 'r' (radius) value and our 'θ' (theta) value, which are essential for defining the region we're integrating over. It's often helpful to sketch the region if possible! I actually calculate the integral in parts 2 and 3, so stay tuned for those!
What You Will Learn:
How to convert Cartesian coordinates to polar coordinates for double integrals.
Setting up the limits of integration based on 'r' and 'θ'.
Understanding the geometry of the region of integration.
Tips for sketching the region to simplify problem setup.
Preparing for the actual calculation of the integral in subsequent videos.
📚 Check out my book: 1001 Calculus Problems for Dummies for more practice!
👍 **If you find this video helpful, please like, share, and subscribe for more math tutorials!
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#Calculus #DoubleIntegrals #PolarCoordinates #MultivariableCalculus #MathTutorial #PatrickJMT #VolumeCalculations #Integration #Mathematics
This video was made 9 years ago and it's coming tremendously useful to me today. Thank you Sir for what you do.
@@mannb1023 "Are the videos from 2008 not helping, well we got the solution for you"
Well actually...
Here 14 years later lol
15 years now,still helping
This comment is over six years old and here it helped me today
16 years check
You my friend have helped thousands of science/engineering/finance students understand Calculus. With any luck, much fortune will come your way.
15yrs ago and it’s helping me now
I owe my engineering degree to you sir...well at least half of it lol
+Asad Khan yes. and no. uni. i took a lot of math courses and his videos helped a lot. i finished all my maths which took up the first two years of school.
@@carlhenry515 How are you now? I felt like I need to know the story of an ex student of Patrick sir :3
@@RoronoaZoro-mw7wu I think you mean Sir Patrick the 3rd
Damn 9 years ago 😮 you probably have a family by now
I will be there no matter what 🏇🏇
@@Salamanca-joro lol
Just passing
you're the secret superhero of everyone's math exams :P
This now 15 year old video is still invaluable today. Thank you for making this, and I hope whoever made this is doing well!
The r term in the equation at 2:33 on the video, comes from substituting x^2 + y^2 = r^2 on the left hand side. And on the right hand side, y/r = sin(theta) so y = r x sin(theta), which is substituted on the right hand side.
You just cleared this concept for me in 2 mins which my uni couldn't in 2 full days of classes....
thanks a bunch mate!!!!
Patrick, dude, I'm completely sure that God's got a special place reserved just for you. You are awesome and from the bottom of my heart: Thank You.
And after 13 years your video helps me to pass math exam , God bless you!
Partick u've saved me, jusr have an exam 2morrow n from 5 days i couldn't understand this matter, thank u dude keep doing this u'll save thousands student's life..thanks from an Angolan in Poland :)
@Iamlaughingman well, the video is not
about graphing polar curves - i am
assuming one knows it at this point
yes, a good graph goes a long ways in helping...
out of all the people that submit videos with lectures, you are my favorite. thank you so much for sharing your knowledge and expertise.!
man, your channel is just incredible .. i am giving up MIT classes and joying PJMT lololololol
good luck on the test esther!
i understood the whole thing in just 5 mins .... where my lecturer took 2 hours and i get nothing ....
yeah this happens to us all
that's why I spend lecture on 9gag and watch these videos when it comes to hw.
11 years ago yet better than professors in 2021
I'M GONNA PASS MY QUIZ BECAUSE OF YOU
thanx patrick..i`ll be sitting for my math exam and this vid has been really helpful..
When going to polar coordinates it is required.
For example, we need an infinitesimal change in the area of the shape. To do this, we multiply a small change in r, by a small change in arc length. This arc length is equal to r*theta, coming from basic trig, and the infinitesimal arc length would be r * dtheta. So when changing to polar coordinates it becomes r dr dtheta.
2021 and you're here saving me
5 stars, you saved me a bunch of wasted time off my 17 hour schedule!
I'm skipping my calculus class to learn what my professor teaches in 1.5 hours by watching this 5 minute video.
@Iamlaughingman he showed where the circle was... you can see the regions. and its in the first and second quadrant. And those quadrants are defined by 0-pi.
You are much better at explaining this stuff than I am, I pretty much destroyed my friends understanding of the material in the process of "helping" him, so I referred him here.
Really?
@patrickJMT well just looking at the circle one can see that since its not at the origin it cant be 0 to 2pi, its above the x axis and that region is 0 to pi
i wish that make ur videos as the book titles and organize them the same to book too :)
as a lessons .. thank you are a great man .. very helpfull ..
"thumbs up"if u think PATRICKJMT is the best tutor:)
Patrick makes math to be easier and simple!
are you single? i could listen to your soothing voice for centuries.
Why is theta between 0 and pi and not between 0 and 2pi if it's a full circle?
It's because in this case, the circle never drops below the x-axis, and r is always greater than zero. If you were continue this graph beyond theta= pi with a positive r, part of the graph would have to be below the x-axis. How it is now, theta moves from zero to pi (positive x-axis to negative x-axis), while the length of r goes from zero to 2sin(theta) and back to zero.
OGP
This is not clear. Can you give links or elaborate on that? First, what coordinate system are you using? It looks like you are using Ortogonal system (with axis - theta and r) and not polar coordinate system to draw the circle. In orthogonal, it should be half a circle - theta goes from zero to (pi). "the circle never goes down" - I see but you have a full circle, not a half circle. In a full circle you always move theta between 0 and 2pi (0 degrees and 360 degrees)
Wow, you are so much better then my professor.
thanks back : )
Hi Pat, I had a question: to fully understand the math in this video I need to know just trig functions, the pythagorean theorem, iterated integrals, double integrals over general regions, general differentiation and general integration (and algebra of course) ?
@JimCharlesTroy ha, happy i could help you : )
It's called the Jacobian. It comes from the change in dxdy to drdtheta. A matrix with partial derivatives is used to determine it.
Very Helpful!
Glad you think so!
I think he did that because the standard eqn. of a circle is [(x-h)^2]+[(y-k)^2]=R^2. Where h and k represent the center point as (h,k).
Why is the bounds for theta 0 to Pi ............. and not 0 to 2 Pi...................I thought how it covers the whole circle we are suppose to use 2 Pi
+Lucas Fernandes The whole circle is above the X axis, which mean you cover the whole circle going from 0 to pi.... Pi to 2pi is under the X axis
check around 2:09, you'll see that he shows this with his pen, because the angle is measured from (0,0) it only takes half a rotation
Lucas Fernandes The circle is above the initial line and in the 1st and 2nd quadrant.
i was reading through comments and noticed yours. Didnt even notice what you pointed out, but this thread helped me learn a pretty important concept i missed learning this. Thanks!
How did you get theta? Why is it from 0 to pi
Why did you opt for a (x-0) and not (x+0) on the left parenthesis? Is it a must that you always put a minus sign?
Thank you sir..
Love from india 😊🇮🇳
May I ask something? Since the radius is 0
This is honestly so helpful. Thank you so much.
Real life saver when teachers dont teach you the materials you need. Read the damn book they say, rage!
this video is why i love youtube
I like it!! It is very useful.
15 years after uploding this vid lol
Great Help!! I hope you were my professor
how would you change the order of integration in polar coordinates?
I love ur tutorials!!
but i got some questions like how do u know what is theta's bounds.. for this one the graph is a simple circle so it's obvious here but if it's other shapes like one with petals... how do u find that out? is the only way to find out is from graph??
Thank you so much!
Thank you, over a decade since.
what is the name of the therome that states the beginig statement
Numbers and alphabets never looked so gosh darn good in 1080p T.T
Please I still don’t get why the r is starting from 0
for a full circle it'S theta between 0 and 2pi , isn't it ?
You saved my life, I love you!
can u do a problem on solving volume for polar coordinates using multi-variable calculus
What would be the limits of integration if polar co-ordinates are not used and this region is integrated in cartesian co-ordinates...Please help..
But the boundary circle lies from 0 to pi. Note that nothing lies 'below' the polar axis, which would correspond from pi to 2pi.
excellent i was looking for this
it is part of the formula! r dr gets added
@patrickJMT So if the entire circle was in the first quadrant, it would be integrated from 0 to pi/4?
you're my hero, i love you
thanks!
Im confused. Since it's a circle, wouldnt it be:
0
How comes you multiplied by an extra 'r' that you said is from the formula. I managed to arrive at the same point as you did, up until the point where you multiplied '12 - 6x -4y' by 'r'
Please can someone explain this.
Thank you
can you explain why theta is between 0 and pi?
isn't a full circle 0 to 2*pi?
This is the most confusing part for me
i believe its because the shaded region is between 0 and pi
Is this 3rd semester of Calculus or Multivariable Calculus?
I was thinking why does R Vary, because it seems to have a constant radius.... However looking at it again, If you take R from the origin, It does vary.... Am i correct saying this? Just thought i should point that out to anyone struggling like i was :S. Not due to explanation due to my little knowledge in Polar Co-ordinates
Pure as we begin
At the beginning he showed a formula for the double integral in polar coordinates where r was a function of theta. Is it also possible to have theta as a function of r, so the r boundary is constants but the theta boundary is a function?
its in radians. Given the function is above the x axis all values are between 0 and 180 degrees. Comprende?
Hey Patrick, does it mater it you integrate with respect to dr of dtheta first. If yes how would I switch the limits of integration. I have some nasty polar curves with gross r's and trig functions and it would be easier to integrate with respect to dtheta first.
Where did the extra "r" come from ?!
Jacobian
thanks helen! (i like that name!)
all around the circle is 360 degrees (2pi), why only Pi? . Secondly what happened to the Z in the equation 6x+4y+z=12?
Its only to pi because we are looking at the disk as a whole on the (r,theta) plane. The equation was put in terms to solve for z, becoming z=12-6x-4y, because were are looking for that area. This is probably not helpful anymore, but just in case anyone else wondered the same thing.
@@lailahasan7615 it was, ty =]
There is someting i cannot get. the R is centered at 1 , not origin. Should not the bounderies of R ; 1 to 2sin(theta)
Nah, r IS 1 AND 2sin(theta).
why did i just reply to a post from 2 years ago? the guys probably an aerospace engineer by now
yay thanks patrick!!!
you're really gr8!!
Thank you! I shall do that now
This helped a lot. Thanks!
is there a more concrete way to find the limits for theta?
thank you so much❤
I passed Calc 1 and 2 thanks to you hahaha
patrickJämeränTaival
Aren't you actually finding the volume under the plane using polar coordinates and not the "area" in the title? I did enjoy your explanations.
how did you figure out the bounds of theta?
But a circle is 2pi, why can't it be greater than pi? please explain in more detail. Thank you!
How is the first limit for r zero? I got how the upper limit is 2sin(theta), but I don't get how the lower limit is 0
How does the body, whose volume we try to find out, look like?
look up 'stretch factor' and it should explain what you need to know ;)
question , why in the first example it is pi and not 2pi?
No..! The circle here is covered completely within the first two quadrants.Hence,from 0 to 2pi.
Had it been a circle with centre (0,0),it would be from 0-->2pi !
Vishwanath V.Pai
Ah, thank you.
how did you know r is between 0 and 2sintheta? i understand the 2sintheta but not the zero
Great teaching !
is there a video explaining why dydx turns into rdrdθ ?
thanks sir it really helped
As usual you are a great help!
Helen