Web7 hours ago · Expert Answer. Transcribed image text: Find the area of the surface formed by revolving the circle r = f (θ)cosθ about the line θ = π/2 (HINT: Area of a Surface of Revolution about the line θ = 2π : S = 2π∫ αβ f (θ)cosθ [f (θ)]2 +[f ′(θ)]2dθ.) WebSep 7, 2024 · Now that we have sketched a polar rectangular region, let us demonstrate how to evaluate a double integral over this region by using polar coordinates. Example 15.3.1B: Evaluating a Double Integral over a Polar Rectangular Region. Evaluate the integral ∬R3xdA over the region R = {(r, θ) 1 ≤ r ≤ 2, 0 ≤ θ ≤ π}.
NEED HELP ASAP, WILL GIVE BRAINLIEST What is the polar form …
WebAbout Press Copyright Contact us Creators Advertise Developers Terms Privacy Policy & Safety How YouTube works Test new features NFL Sunday Ticket Press Copyright ... WebFind the area inside of the bigger loop of r=1+2cos(theta) but outside of the smaller loop.⭐️Please subscribe for more math content!☀️support bprp on Patreon... iron chicken clangers
Cos Theta 2 - BRAINGITH
WebCalculus. Graph r=2cos (theta) r = 2cos (θ) r = 2 cos ( θ) Using the formula r = acos(θ) r = a cos ( θ) or r = asin(θ) r = a sin ( θ), graph the circle. r = 2cos(θ) r = 2 cos ( θ) WebDifferentiate both w.r.t. θ We get: d θ d x = − 3 a cos 2 θ sin θ and d θ d y = 3 a sin 2 θ cos θ Thus d x d y = d θ d x d θ d y = − 3 a cos 2 θ sin θ 3 a sin 2 θ cos θ = − tan θ WebWhat is the polar form of this equation? \[ \begin{array}{c} (x+2)^{2}+y^{2}=4 \\ r^{2}=4 r \cos \theta \\ r^{2}=4 r \sin \theta \\ r(r+4 \cos \theta)=0 \\ r(r+4 \sin \theta)=0 \end{array} \] … port numbers is used by telnet