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- If the electric intensity in the vacuum E = Eocosθar - Eosinθaθ is the divergence of the electric field strength.and find the curlFind the area of the inner loop of the curve r= 1-2cos θUse Green’s theorem to evaluate ∮C(ye2xy−5y)dx+ (xe2xy−2x)dy, where Cis the counterclockwise oriented boundary curve of the square with vertices at(0,0), (0,1), (1,0), and (1.1).
- Find the EXACT area inside the closed loop produced from the following parametric curve. Show ALL work. x= 63t-7t^3 y= 1+t^2Find the area enclosed by one loop of this polar curve: r=3sqrt(cos2theta) from 0 to 2pi using the formula A=1/2 integral from 0 to 2pi (r)^2 for parametric curve.Show that the path given by r(t) = (cos t,cos(2t), sint) intersects the xy-plane infinitely many times, but the underlying space curve intersects the xy-plane only twice.
- Find the areas of the regions Inside one loop of the lemniscate r2 = 4 sin 2θConsider I = ∫CF⋅dr, where (img17) is a conservative vector field and curve C is parameterized by:α (t): = ((2 − cos (5t)) cost, (2 − cos (5t)) synt, sin (5t)) with 0≤t≤π. We have that the value of I is equal to: (img18)Find the area outside the curve r=3+2cos(-) and inside the curve r=3- 3 cos(-)
- But the definition of subharmonic is -laplace(v) less than or equal 0 in U. How could we obtain that?Find the arclength of the path R=pie cos3t I + 4 t^(3/2) j + pie sin3t k on interval between and including 0 and 2.Suppose that U is a solution to the Laplace equation in the disk Ω = {r ≤ 1} andthat U(1, θ) = 5 − sin2θ.(i) Without finding the solution to the equation, compute the value of U at theorigin – i.e. at r = 0.(ii) Without finding the solution to the equation, determine the location of themaxima and minima of U in Ω.(Hint: sin2θ =(1−cos 2θ)/2.)