If C is a simple closed curve in the plane enclosing the region R, then we can use Green's Theorem to show that the area of R is 1/2∫C x dy - y dx. (a) Find the area of the region enclosed by the ellipse r(t) = (acos(t))i + (bsin(t))j for 0 ≤ t ≤ 2π. (b) Find the area of the region enclosed by the astroid r(t) = (cos^3(t))i + (sin^3(t))j for 0 ≤ t ≤ 2π.
Added by Jaime F.
Step 1
Then dx = -asin(t)dt and dy = bcos(t)dt. Substituting these into the formula for the area, we get: Area = 1/2 ∫C x dy - y dx = 1/2 ∫_0^2π (acos(t) * bcos(t)dt - bsin(t) * -asin(t)dt) = 1/2 ∫_0^2π (abcos^2(t) + absin^2(t))dt. Since cos^2(t) + sin^2(t) = 1, we can Show more…
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Calculating Area with Green's Theorem If a simple closed curve C in the plane and the region $R$ it encloses satisfy the hypotheses of Green's Theorem, the area of $R$ is given by Green's Theorem Area Formula Area of $R=\frac{1}{2} \oint_{c} x d y-y d x$ The reason is that, by Equation (4) run backward, Area of $R=\iint_{k} d y d x=\iint_{k}\left(\frac{1}{2}+\frac{1}{2}\right) d y d x$ $$=\oint_{c} \frac{1}{2} x d y-\frac{1}{2} y d x$$ Use the Green's Theorem area formula given above to find the areas of the regions enclosed by the curves in Exercises $31-34$ The astroid $\mathbf{r}(t)=\left(\cos ^{3} t\right) \mathbf{i}+\left(\sin ^{3} t\right) \mathbf{j}, \quad \mathbf{0} \leq t \leq 2 \pi$
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