Area of Polar Curve r=1+2cos(theta)
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- Опубликовано: 11 сен 2024
- Find the area inside of the bigger loop of r=1+2cos(theta) but outside of the smaller loop.
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u sir just saved my life, i surely do appreciate this considering how i struggle with polar concepts
As to why the integrant is (1/2)*r^2: When 'r' is a constant you get a perfect circle with theta going from 0 to 2Pi. This is the equivalent of an horizontal line in the traditional rectangular coordinates. In rectangular coordinates, when you define step functions (from above and below) you get an area of a rectangle for each partition. In polar coordinates when you define a step function you get the area of a circular segment which is equal (1/2) * (theta_k - theta_k-1) * r^2. Where 'r' stand for radius and 'theta_k - theta_k-1' is the angle size of a given partition 'k' in radians.
It's very clear lecture for every students who studying polar coordinates..... thank you
3D geometry please please please please please please
Why is the area for a polar graph the integral of 1/2*r^2?
The Jacobian or density correction factor required for integrating correctly over sets in non Cartesian coords can be calculated using the defn of the change of variables. Calculation gives dx dy = r dr d(-). From here, if we have r ranging from 0 to r((-)), then this becomes dx dy = ½r((-))² d(-) by integrating over r. Here, (-)=theta
The strip area for a polar graph is a form of a triangular shape which area is (0 .5absin(theta)) that's why the integral area is 0.5r^2d(theta). a and b is just r and sin(theta) becomes d(theta) because it's super small we get rid of "sin". ( no good in english sorry.)
3 different definitions and I still dont know
@@supersayinsauce1788 Biggest fallacies of smart people teaching maths...they usually state it in terms that *they* understand, but not necessarily comprehensible to someone trying to learn it for the first time.
The straightforward, analytical answer is that the area of a polar graph is always a double integral, the outer integral is the same as shown in the video, but the inner integral has bounds 0 to r, integrating over dr, times the Jacobian which is just the determinant of a matrix, and is r for polar co ordinates. Computing this inner integral (with those bounds) will always give you r^2 / 2.
You're not alone. Important step was glossed over and was stated as the "formula of the area." The area formula in polar coordinates should have been explained or derived.
If you have a calculus text book, go to the index and look for "area under polar coordinates" for the explanation of the formula. Good luck. Stay safe and in good health.
Now do... "Area of Polar Curve, Calculus 3" :)
I don't understand why for the smaller semicircle section the left end is 2pi/3? Isn't that point on the x axis? Wouldn't you expect it to be 0, pi or 2pi?
at 2pi/3 r = 0 and the smallar circle starts to grow and at 4pi/3 it is complete. if you draw the equation from theta = 0 to 2pi you will see from 2pi/3 to 4pi/3 the smaller circle is complete. taking the integration limits as 2pi/3 to 4pi/3 will result in the smaller circle's area.
Could the final answer be 3π? First, we can cancel out the halfs since we have a factor of 2, and then, we can multiply the expression out to get 1 + 4cosθ + 4cos^2θ. Integrating this, we get 3θ + 4sinθ + sin2θ. If we evaluate this from 0 to π, we get the answer of 3π - everything else cancels out, given that whole-numbered multiples of π are the zeroes of the sine function.
Ben pi +3×3½ buldum. Ama kontrol etmedi.
I got pi only tho :”/
very good guy, congratulations, your channel is awesome
Plzz integrate
ln(ln(ln(lnx)
: )
Neigh!
Horseshoe.
Maybe ln(ln(.......(x))))) in general :D
@@blackpenredpen Do you have a video showing how to derive the formula for polar curves though? That's an important part. Or is there no way to derive it?
Great explanation, clear and concise, by the way, could you comment on how to find the interval of theta that the graph is trace out exactly once? thanks appreciated it
Thank you sooooooooooo much for making my concept
Gracias, te amamos!
Your videos is very helpful to students
: )
Please do rotational geometry! Please! I will die a peaceful man then!
Such as what tho?
Like ‘If a prism is rotated...’
Hablo español y te entendí, que grande 👏
realllyyyyyyyyyyy lol, now I understand !!!! thanks BPRP!!!!!!
Area of green portion is negative.so can i say u have to add not subtract it to get exact value?
Thats true, actually the master is wrong here
it should be just one int. from 0 to Pi
hopefully he will change it
@@hugoirarrazavalarteaga9879 actually the answer is right, you are already going from a higher bound to the lower
The ans is π + 3 √3
Derek Poon you’re off by a little bit
@@PLANET-EATER mm no i put the integal into my calculator and got a decimal value equal to (pi + 3sqrt(3))
i still dont understand how you get the Pi for the integral
Pi is in the limits of integration. Very common for polar integrals, because we want a broad sweep across all possible values of theta, from 0 to 2*pi. After theta exceeds 2*pi, it becomes a redundant part of the function, so polar domains are commonly limited from theta = 0 to 2*pi.
BPRP, you legend... Thank you so much.
Prove that ln(x) is the only integral of 1/x and no other polynomial :D
Jan V. 1/x is not a finite polynomial. Anyway, your thesis is false. Ln(-x) is also an integral of 1/x, as is Ln(x) + 1 and Ln(-x) - 1.
I thought the integral of (1/x)dx was ln|x|?
@@jmcwd It is. Although that's a half-truth, because the integral of 1/x is actually discontinuous at the vertical asymptote. For most practical applications, it's good enough, and we keep it simple as ln|x| + C.
In the general case, the +C can be different on both sides of x=0, so it is more like ln|x| + C + D*u(x), where u(x) is the unit step function, and D is another arbitrary constant. There are very few (if any) applications where this matters.
If we extend the domain to complex numbers, the integral of 1/z dz is the complex log of z + C. The complex log of z is the natural log of the magnitude of z, plus i*(theta + 2*pi*k), where k is any integer, and theta is the angle of the complex number z. So it becomes:
integral 1/z dz = ln(|z|) + i*(arg(z) + 2*pi*k) + C. We have two arbitrary constants in this integral, the k because there are multiple branches of natural log, and C because it is an indefinite integral.
You can do a limit proof on the power rule, to show why the integral of 1/x dx cannot be a standard polynomial, and instead is natural log.
Let the exponent be -1 + h, such that the function is x^(-1 + h). The integral of a power function in general, of exponent n, is 1/(n + 1)*x^(n) + C.
Substitute n = -1 + h:
1/(-1 + h + 1)*x^h + C
Simplify:
1/h*x^h + C
Use L'H's rule to take the limit as h approaches zero:
d/dh x^h = d/dh e^(ln(x)*h) = ln(x)*e^h. Evaluate at h=0, gives us ln(x)
d/dh h = 1
Compiling the limit gives us:
limit as h->0 of 1/h*x^h = ln(x)
And since +C is independent of h, we can simply add +C
Result:
integral 1/x dx = ln(x) + C
i'll leave it as an exercise to you, to show that applying absolute value bars to the x, makes this also valid for negative x.
Can you please elaborate how to find radius of curvature of any random curve(e.g. ellipse,hyperbola,etc) if don't know its differential form
Nice but you lost me where it's the direction of positive x-axis and still wrote "r" as minus one "-1" mmm???
Why do you integrate 1/2 r^2 dTheta ?... Shouldn't it be 1/2 r dTheta? i.e. we're cutting the graph into tiny triangles with base equal to r and height equal to dTheta.
bilz0r We’re integrating the area which means we need a double integral:
int(0,r, int(0,θ1, r dθ)dr)
From Fubini’s theorem for double integrals:
int(0,θ1, int(0,r, r dr)dθ)
And we get:
int(0, θ1, (1/2)(r^2) dθ)
Where r is a function of θ
Where the notation
int(a,b, f(x) dx)
Should be read as “the integral from a to b of f(x)dx.”
I'm just an idiot. I thought the height was equal to dTheta, but obviously the height is dTheta*R... so the area of each triangle is 1/2*r*r*dTheta = 1/2 r^2 dTheta.
bilz0r We’re dividing the curve with tiny wedges, not necessarily triangles. The area of a wedge with radius r is 1/2*r^2*theta, where theta is the angle of the wedge in radians. As theta becomes infinitesimally small you end up with 1/2*r^2*dtheta. This is what you integrate.
gracias , estaba buscando uno en español pero no lo pude encontrar igual se pudo entender muy bien . Muy buena explicación!!
Bro, why are you so amazing!
Found the equation for tilak in India
So you found many types of geometry curve and equation on the forehead mathematical representation of universe on matha (forehead)
Where actual mathematical calculation occurred
Your just pumping out the videos nowadays
The result of the 2nd integral is negative, isn't it? So you end up adding the two areas instead of subtracting. Shouldn't you be adding the two integrals?
he should either change the bounds, or sign
Good observations. Area below the initial line is negative. I think another negative needs to be introduced Infront of the second integral to get the difference instead of the sum
Why is theta2 pi? And what does a negative value of r indicate? I didn't get that part...
おお、なんかむっちゃり難しそうに見えたけど
普通に積分するだけでいいんですね
実数しか出てこんかった🙂
What is hydraulic jack formula in integrated form so we know it's maximum capacity and thresholds limits at quantum physics level
Me encantoooooo.....
2cosθ=-1
cosθ=-1/2
θ=2π/3, -2π/3
Integral from - to + minus integral from + to - of r^2 (or really you can do it the other way around, you’ll just get a negative version of the answer)
edit: forgot the 1/2 xd
Do more videos about this concept.
Words cannot express how much I love you. Please marry me
thank you
Where did the formula for the integral at 4:30 come from?
Think about it this way. When we use Riemann sums to establish integrals in general, we're slicing the function like it is a loaf of bread. We use the value of the function to tell us the size of each slice of bread, and then add up the slices to tell us the size of the total loaf.
In polar coordinates, instead of slicing bread, we are slicing a pizza. Imagine a homemade pizza that isn't perfectly round, but has a radius that is a function of its theta-position. We find the area of each slice, by approximating it as a circular sector, and then add up the area of each slice. The area of a circular sector (a pizza slice) is 1/2*r^2*theta, where theta is its angle in radians and r is the radius. The infinitesimal pizza slice, has angle dtheta instead of theta, so the area element has area 1/2*r^2*dtheta.
I cannot understand how come that r=-1 because by definition r=sqrt (x^2+y^2) so I don't get it.More challenging how do u define that region? r between ? and theta between?Is it possible that when u have
2/3pi
Federico Espil the points of the curve are defined by
(x,y)=(r cos(θ),r sin(θ)) and r = f(θ) so it can be negative, and the def is not the distance to origin (even if abs(r) give it) here. it is like multiplicating a complex Z of U={complex z such that |z|=1}, taking a real r value and say |rZ|=r
Sed but the definition of r I read in lots of pdf even Wikipedia is the distance from the point to the origin Whatever it is the definition it is troublesome to define that area (not the curve) in polar system because of this problem
No, i talk about the définition of r not "the radius", just r, this is just a function of θ.
You saved my life
perfect video
Sir can you do the same for 1 +cos2theta
Thanks!
omg THANK YOU
quick answer:
r = 1+cos(theta)
integrate from 0 to 2pi/3 of (1+cos(theta))^2 - integrate from 2pi/3 to pi of (1+cos(theta))^2
Amazing!
Hi, can you please find the area of inner loop?
Any videos on how to derive this formual for this cardiod shape?
why the second integral from 2π /3 to π an not 2π ?
At 2pi, you’re just going to go back at theta = 0, and we don’t want that. Look at the graph, and then trace the steps where it’s going from. It will help remember where this is going.
Does the answer of the area approx. to 8.33?
life saver
I think the method you use to determine theta for cos(theta) = -1/2 is flawed. It omits the second solution, and while in this case it works out, that's not always the case. Also if you interpret -1/2 as 1/(-2) (the different but equivalent interpretation of the fraction) you have to do some mental gymnastics to end up with the correct triangle for the omitted solution.
r is always positive. This is called the reference triangle, it's a precalc method. : )
then you're just omitting the second solution
tbh I'd still just draw a unit circle
From the symmetry of the problem, it’s obvious the angle sought after is on the interval (0, π]. The second solution proposed is not relevant.
Even if it were, interpreting -1/2 as 1/(-2) does not remedy the problem. More generally, we can still arrive at the proposed second solution using reference triangles by taking the r into the third quadrant.
For more, consult:
www.nr.edu/chalmeta/271DE/Section%209.4.pdf
@@Jhopsssss to construct the triangle for 1/2 and then take the full negative. The issue is you end up with -1 for x again even though we avoided that.
The best reasonable explanation for why to do this is to just use the unit circle.
Sticking to the method, the best method would be to lock in x=-1 and then draw R either direction to complete the triangle, and THEN reject any solution that you don't need.
@@BigDBrian, in this particular example of cosθ = -1/2, the solutions on the interval [0, 2π) are θ = 2π/3 and θ = 4π/3. x must be -1 in order to arrive at these solutions because the vector sweeping out these angles occupy the second and third quadrants. Reversing the direction of r in either of these cases lands the vector in the first or fourth.
The reference triangle method begins by plotting x = -1 (a vertical line on the Cartesian plane). Then, we can draw r = 2 (a circle of radius 2 centered at the origin) and find the two points at which they intersect. We then draw a vector to each of those points and find that the angles they sweep out are 2π/3 and 4π/3.
To address your concern of chronology: at this point, we are free to discard the 4π/3 solution.
Please note that we cannot simply draw a horizontal line segment of length 1 and then draw a diagonal line segment of length 2 in any direction and expect any useful information to fall out from it. We would get an family of triangles. The reference triangles are defined such that the projection of r onto the x-axis is equal to x (i.e. x = -1 in this case).
thank you :)
Derive the formula without double integral
I tried with formula for parametric curves
interestingly this polar function is nearly approximated (on the right side) with a circle of radius of sqrt(pi)
i.imgur.com/GZ5sokp.jpg
Why did you square that
i really love u.
Intigration( x^2 e^x^2 dx)
any limits?
this curve actually does not intersect the origin it is very close but if you graph it on desmos and zoom in to the "intersection" you'll see it isn't one
nice
Great 👍🏻
God bless you
Not the origin. The pole.
saxbend (0,0) is the origin by definition
@@angelmendez-rivera351 only in cartesian coordinates. Polar coordinates are based on modulus and argument. Modulus being represented by distance from the pole.
saxbend No, it represents the origin in all coordinate systems. It is defined as the origin in all coordinate systems. And I know how spherical coordinates work. I have a degree.
Do you dream in numbers?
If we integrate or differentiate
3x^2+2x+1 = 0
Then what are applications of it?
If we integrate it, we can find the area under it's curve, if we differentiate it we can find the slope of a tangent line at any point, those are the simple ones there are many more applications
Mohna Khan Simply put, you can calculate the area under the curve, or instantaneous change at any point in time.
@@georgefan2977 actually, it's impossible to know the instantaneous rate of change at a single point, but it does give a back a closely approximated value. For example, the derivative of x^2 is 2x. That means the slope of a tangent line would tend to the value of 2 times the value you chose.
Pie is the answer
I got pi. :D
love from indi
Szacun🕊✋🌷salut!!!!🎺👒
Limit should be 0 to pi
Great
I got (pi+3sqrt3) as a final answer
Are you studying at hkust
integral of y×dx is area
In rectangular coordinates, yes. We're slicing a loaf of bread, and the height of each slice is y, and the thickness of each slice is dx. Integrating y dx gives us the area of the loaf. In 3-D, it becomes a double integral, where we're slicing a corn bread, where the width is dx, the length is dy, and the height is z, so we're integrating twice, the value of z*dx*dy.
In polar coordinates, we're slicing a pizza. Each slice has an angle of dtheta, and a radius of r. The area of each slice is 1/2*r^2*dtheta.
Thank Q very much ....
Where do you live? China or somewhere else?
He lives in the US
Quisiera que usted fuera mi catedrático de cálculo, no uno todo pendejo :)
😅😅😅This polar stuff is confussing
Did I watch this before?
I think so. : )
😂 and I think I have seen that before.
What if I have equation 1/2 + cos(theta). Is it the same as 1+2cos(theta)??
Hey BPRP. Could you do the integral of x^2/sqrt (4+x^2) dx using hyperbolic substitution?
Great, That's the same int that I wasn't able to solve in my math test without hyperbolic sub, BPRP could you solve it with hypsub and with algebraic method?
This is a CARDIOID..
pi + 5sqrt(3) is what I got can anyone verify?
Can any one say how to form this curve in desmos app
Karthick Raja Type “r=1+2cos(theta)”
@@AnnoyingMiner10 yess i made mistake i put x instead of r i.e., i mixed Cartesian coordinates and polar coordinates in a same jar :-)
Karthick Raja oof That’s alright!
I was wondering what happened to this video.
R maior ou igual zero!!!!
green pen entered the exercise, unsubscribe
First!
thanks!