# Green's Theorem help

#### tg25007

##### New member
8. Evaluate $$\displaystyle \int F \cdot dr$$ both directly and by Green’s Theorem. The vector field and the region D is the upper half of a disk of radius a: $$\displaystyle 0 \leq x^2 + y^2 \leq a^2$$ , $$\displaystyle y$$ . 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>

$$\displaystyle \int F \cdot dr = \int <a^2, asint> \cdot <0, acost>$$
$$\displaystyle = \int^{pi}_0 a^2sintcost dt$$
$$\displaystyle = \frac{1}{2}a^2sin^2t$$
$$\displaystyle = 0$$

For Green's Theorem, I should get the same number, but I got $$\displaystyle -\frac{4}{3}a^3$$ and I have no idea where I went wrong.

$$\displaystyle \int \int (\frac{d}{dx}(y) - \frac{d}{dy}(x^2 + y^2)) dA$$
$$\displaystyle = \int \int -2y dA$$

Then I polarized the coordinates where y = rsin(theta)

$$\displaystyle \int^{\pi}_0 \int^a_0 -2rsin(\theta)r dr d(\theta)$$
$$\displaystyle = \int^{\pi}_0 -\frac{2}{3}r^3sin(\theta)$$ 0 to a
$$\displaystyle = \int^{\pi}_0 -\frac{2}{3}a^3sin(\theta) d(\theta)$$
$$\displaystyle = \frac{2}{3}a^3cos(\theta)$$ 0 to pi
$$\displaystyle = -\frac{4}{3}a^3$$

#### Fernando Revilla

##### Well-known member
MHB Math Helper
8. The vector field and the region D is the upper half of a disk of radius a: $$\displaystyle 0 \leq x^2 + y^2 \leq a^2$$ , $$\displaystyle y$$ . The curve C is the boundary of D and oriented counter-clock wise.
Hello tg25007, welcome to MHB. I'd like to help you, but you don't mention $\vec{F}.$

#### Opalg

##### MHB Oldtimer
Staff member
8. Evaluate $$\displaystyle \int F \cdot dr$$ both directly and by Green’s Theorem. The vector field and the region D is the upper half of a disk of radius a: $$\displaystyle 0 \leq x^2 + y^2 \leq a^2$$ , $$\displaystyle y$$ . The curve C is the boundary of D and oriented counter-clock wise.

I got an answer of 0 using the direct way:
Let $x = a\cos t, y = a\sin t$
$F(t) = <a^2, a\sin t>$
$F'(t) = <0, a\cos t>$

$$\displaystyle \int F \cdot dr = \int <a^2, a\sin t> \cdot <0, a\cos t>$$
$$\displaystyle = \int^{\pi}_0 a^2\sin t\cos t\, dt$$
$$\displaystyle = \frac{1}{2}a^2\sin^2t$$
$$\displaystyle = 0$$

For Green's Theorem, I should get the same number, but I got $$\displaystyle -\frac{4}{3}a^3$$ and I have no idea where I went wrong.

$$\displaystyle \int \int (\frac{d}{dx}(y) - \frac{d}{dy}(x^2 + y^2)) dA$$
$$\displaystyle = \int \int -2y dA$$

Then I polarized the coordinates where y = rsin(theta)

$$\displaystyle \int^{\pi}_0 \int^a_0 -2r\sin(\theta)r\, dr d(\theta)$$
$$\displaystyle = \int^{\pi}_0 -\frac{2}{3}r^3\sin(\theta)$$ 0 to a
$$\displaystyle = \int^{\pi}_0 -\frac{2}{3}a^3\sin(\theta)\, d(\theta)$$
$$\displaystyle = \frac{2}{3}a^3\cos(\theta)$$ 0 to pi
$$\displaystyle = -\frac{4}{3}a^3$$
You have the correct result, $-\frac43a^3$, with the Green's theorem calculation, but you have messed up on the $\int F\cdot dr$ part. For a start, you need to have two integrals to go round the contour $C$, first round the semicircle and then along the $x$-axis. Next, you need to work out $\int F\cdot dr$ for each of those pieces of the contour. On the semicircle, you correctly have $F=(a^2,a\sin t).$ You don't need to differentiate $F$, but you should have $r = (a\cos t, a\sin t)$ and so $dr = (-a\sin t, a\cos t)dt.$ Take the dot product of that with $F$ and integrate from $0$ to $\pi.$ For the part of the contour that goes along the axis, you have $F = (t^2,0)$ and $r = (t,0)$ (with $t$ going from $-a$ to $a$). Add the two integrals together and you should get $-\frac43a^3$.

LaTeX tip: put a backslash before cos and sin, and a space after them, if you want them to display properly.

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#### Ackbach

##### Indicium Physicus
Staff member
LaTeX tip: put a backslash before cos and sin, and a space after them, if you want them to display properly.
Another $\LaTeX$ tip: for iterated integrals, it looks better if the differentials are separated by a
Code:
\,
space thus:
$$\int_{a}^{b} \int_{c}^{d}x^{2} \,dx \,dy.$$