tg25007
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8. Evaluate $$\int F \cdot dr$$ both directly and by Green’s Theorem. The vector field and the region D is the upper half of a disk of radius a: $$0 \leq x^2 + y^2 \leq a^2$$ , $$y$$ . The curve C is the boundary of D and oriented counter-clock wise.
I got an answer of 0 using the direct way:
Let x = acost, y = asint
F(t) = <a^2, asint>
F'(t) = <0, acost>
$$\int F \cdot dr = \int <a^2, asint> \cdot <0, acost>$$
$$= \int^{pi}_0 a^2sintcost dt$$
$$= \frac{1}{2}a^2sin^2t$$
$$= 0$$
For Green's Theorem, I should get the same number, but I got $$-\frac{4}{3}a^3$$ and I have no idea where I went wrong.
$$\int \int (\frac{d}{dx}(y) - \frac{d}{dy}(x^2 + y^2)) dA$$
$$= \int \int -2y dA$$
Then I polarized the coordinates where y = rsin(theta)
$$\int^{\pi}_0 \int^a_0 -2rsin(\theta)r dr d(\theta)$$
$$= \int^{\pi}_0 -\frac{2}{3}r^3sin(\theta)$$ 0 to a
$$= \int^{\pi}_0 -\frac{2}{3}a^3sin(\theta) d(\theta)$$
$$= \frac{2}{3}a^3cos(\theta) $$ 0 to pi
$$= -\frac{4}{3}a^3$$
I got an answer of 0 using the direct way:
Let x = acost, y = asint
F(t) = <a^2, asint>
F'(t) = <0, acost>
$$\int F \cdot dr = \int <a^2, asint> \cdot <0, acost>$$
$$= \int^{pi}_0 a^2sintcost dt$$
$$= \frac{1}{2}a^2sin^2t$$
$$= 0$$
For Green's Theorem, I should get the same number, but I got $$-\frac{4}{3}a^3$$ and I have no idea where I went wrong.
$$\int \int (\frac{d}{dx}(y) - \frac{d}{dy}(x^2 + y^2)) dA$$
$$= \int \int -2y dA$$
Then I polarized the coordinates where y = rsin(theta)
$$\int^{\pi}_0 \int^a_0 -2rsin(\theta)r dr d(\theta)$$
$$= \int^{\pi}_0 -\frac{2}{3}r^3sin(\theta)$$ 0 to a
$$= \int^{\pi}_0 -\frac{2}{3}a^3sin(\theta) d(\theta)$$
$$= \frac{2}{3}a^3cos(\theta) $$ 0 to pi
$$= -\frac{4}{3}a^3$$