Calc 3: Strokes Therom Orientatio Question

[Tom McCurdy]

Homework Statement

$$F(x,y,z) = (x^2y^3z)i +(sin(xyz))j +(x^2yz)k$$
S is the part of the cone $$y^2=x^2+z^2$$ that lies between the planes y = 0 and y = 3, oriented in the direction of the positive y-axis

My question deals with orientation (see below sec 3)

Homework Equations

$$\int_c F dr = \int\int_S F dS$$

The Attempt at a Solution

Alright so the boundary curve C is the circle $$x^2 + y^2 = 9$$ $$y=3$$

My question is why does r(t) become
$$r(t) = 3sin(t) i+ 3j + 3cos(t)k$$
instead of
$$r(t) = 3cos(t) i+ 3j + 3sin(t)k$$

I am assuming it has something to do with the positive orientation towards positive y axis

When I did the problem I got the exact negitive of what the answer should be

i got $$\frac{-2187}{4}\pi$$ instead of $$\frac{2187}{4}\pi$$

so I was also going to ask if I mess up the cos and sin like i did on this one should I always get the opposite sign of the correct answer or was this just random?

Also should all Strokes theorms problems be positive answers?

 

[benorin]

so I was also going to ask if I mess up the cos and sin like i did on this one should I always get the opposite sign of the correct answer or was this just random?

If you swap sin and cos in your parameterization of the curve C, you will get the negative answer (provided there aren't any other curves to be integrated along) since this effectively reverses the orientation of the curve from C to -C (C traversed backward), this gives a negative answer because

$$\int_{-C}\vec{F}\cdot d\vec{r} = -\int_{C}\vec{F}\cdot d\vec{r}$$.

Also should all Strokes theorms problems be positive answers?

No, absolutely not.