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.?
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.
Ur mom @Chigga Wigga
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.
To compute the area enclosed by the curve defined by the parametric equations, we can use the formula for the area of a closed curve given by Green's theorem:
A = 1/2 ∮ (x dy - y dx)
Let's break down the steps to compute the area:
Step 1: Compute dx and dy in terms of θ.
To find dx, differentiate x(θ) with respect to θ:
dx/dθ = -Asinθ
To find dy, differentiate y(θ) with respect to θ:
dy/dθ = Bcosθ
Step 2: Compute the integrand (x dy - y dx).
Substitute the expressions for dx and dy into the integrand:
(x dy - y dx) = (Acosθ+h) (Bcosθ) - (Bsinθ+k) (-Asinθ)
= ABcos²θ + Ahcosθ + BAsin²θ + Bksinθ
Step 3: Integrate the integrand over the closed curve.
The closed curve is defined by 0≤θ≤2π, so the integral becomes:
A = 1/2 ∮[0,2π] (ABcos²θ + Ahcosθ + BAsin²θ + Bksinθ) dθ
Step 4: Evaluate the integral.
To compute the integral, we need to break it down into four parts and evaluate each part separately:
A = 1/2 [∫(0,2π) ABcos²θ dθ + ∫(0,2π) Ahcosθ dθ + ∫(0,2π) BAsin²θ dθ + ∫(0,2π) Bksinθ dθ]
The first integral is a standard integral that evaluates to πAB. The third integral also evaluates to πAB. The second and fourth integrals evaluate to zero because integrating over a full period of cosine or sine results in zero.
Therefore, the simplified expression for the area A is:
A = πAB + 0 + πAB + 0
A = 2πAB
So, the area enclosed by the curve defined by the parametric equations x(θ) = Acosθ + h and y(θ) = Bsinθ + k, for 0 ≤ θ ≤ 2π, is 2πAB.
Yo it's 2019 tf you answering for? @Bamboozler
eweeee
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