class: center, middle, inverse, title-slide .title[ # Polar Coordinates ] .subtitle[ ## Session 11 ] .author[ ### Alejandro Ucan ] .date[ ### 2024-04-07 ] --- # Goals for the session * Know the definition of polar coordinates. <br/><br/> * Establish the change of coordinates identities. <br/><br/> * Show how the integral is affected in this new coordinate system. --- # Motivation ### Polar coordinates in `\(\mathbb{R}^2.\)` <iframe src="https://www.geogebra.org/calculator/aubzbhsd?embed" width="1200" height="400" allowfullscreen style="border: 1px solid #e4e4e4;border-radius: 4px;" frameborder="0"></iframe> --- # Change of Variables > If we have a point `\((x,y)\)` in `\(\mathbb{R}^2\)` then the following holds: <br/><br/> * `\(r=\sqrt{x^2+y^2}\)` and `\(\tan(\theta)= \frac{y}{x}.\)` <br/><br/><br/> * `\(x=r\cos(\theta)\)` and `\(y=r\sin(\theta)\)` --- ## Express Polar Regions > Given the following regions determine if they are polar regions or not.  --- # The Area Differential in the new system > Compute the area of the region bounded by the `\(x-\)`axis and the curve `\(f(x)=\sqrt{1-x^2}\)` using polar coordinates. --- # Adapting the Area differential From the previous example we have that the area is given by `$$\int\int_R dxdy=\int\int_R rdrd\theta.$$` > Recompute the area using this adapted differential. --- # Example: > Compute the value of the integral of `\(f(x,y)=e^{x^2+y^2}\)` in the positive half of the unit circle. --- # Example: > Compute the value of the integral of `\(f(x,y)=e^{x^2+y^2}\)` in the positve hakf of the unit circle. `$$\int\int_R e^{x^2+y^2}=\int_0^\pi\int_0^1 e^{r^2} rdrd\theta =\int_0^\pi \left[\frac{1}{2}e^{r^2}\right]_0^1 d\theta = \frac{\pi(e-1)}{2}.$$` --- # Practice: > Compute the integral of `\(f(x,y)=x\)` in the first cuadrant of the disk of radius `\(a.\)` --- # Why is it better to calculate the integral with a change of variable? > Using Rectangular coordinates, find the value of the integral `$$\int_0^1\int_0^{\sqrt{1-x^2}}x^2+y^2 dydx.$$` ---  `\(C=\{(r,\theta): 0\leq r\leq 1, 0\leq \theta \leq \frac{\pi}{4}\}.\)` Por lo que la integral es: `$$\int_0^1\int_0^{\pi/4} (r\cos(\theta))^2+(r\sin(\theta))^2 rd\theta dr=\int_0^1\int_0^{\pi/4}r^3d\theta\ dr =\left.\frac{\pi}{4}\frac{r^4}{4}\right|_0^1=\frac{\pi}{4}.$$` --- # Example: > Find the volume of the solid region above the unit circle and below the paraboloid `\(z=9-x^2-y^2.\)` --- # Example > Find the volume of the solid region below `\(z=4\)` and above `\(z=x^2+y^2\)`