the area is just
∫[0,2π] y(t) dx/dt dt
= ∫[0,2π] (Bsint+k)(-Asint) dt
= ∫[0,2π] -ABsin^2(t) - kAsin(t) dt
= (-AB/2)(t-sint cost) + kAcost [0,2π]
= (AB/2 (2π) + kA) - (kA)
= πAB
This is true, of course, since the curves describe the ellipse
(x-h)^2/B^2 + (y-k)^2/A^2 = 1
and the area of an ellipse with semi-axes A and B is πAB
Compute, in terms of A, B, h, and k, the area enclosed by the curve defined by parametric equations. x(θ)=Acosθ+h and y(θ)=Bsinθ+k. 0≤θ≤2π.?
6 answers
eweeee
So these set of equations give an eclipse. Area of any eclipse will be piAB where A and B are coefficients of sine and cosine. The reason why h and k donot affect the area is because the eclipse is merely shifted up or down and right or left by these values and the shape of the eclipse is unaffected.
Yo it's 2019 tf you answering for? @Bamboozler
Ur mom @Chigga Wigga
A little correction to Bamboozler's comment:
Since A and B can also take negative values, the total unsigned area would be: |-piAB|. The minus sign is there when calculations are done.
Since A and B can also take negative values, the total unsigned area would be: |-piAB|. The minus sign is there when calculations are done.