Integral under a closed curve

**Showcase_22** · April 24th 2009, 01:49 AM

Let H be the curve parameterised by $\underline{r}(t):=(sin ~t, cos~t, sin~2t), \ 0 \leq t \leq \pi$ , and let L be the horizontal line from $(0,-1,0)$ to $(0,1,0)$ .

Let $\underline{v}$ be the vector field $\underline{v}(x,y,z):=(2z \ , \ y(x^2+y^2) \ ,\ x)$ .

Question: Evaluate $\int_C \underline{v} \cdot d \underline{r}$ where $C$ is the closed curve formed by joining H and L (You may use either orientation of C).

I can't picture geometrically what this question is asking me to do.

Here's my attempt at a solution:

$\int_C \underline{v} \cdot d \underline{r}=\int_a^b\underline{v}(\underline{r}( t))|\underline{r}'(t)|~dt$ where $a=(0,-1,0)$ and $b=(0,1,0)$ .

$=\int_a^b (2sin(2t), \ sin(2t), \ sin(t)) |r'(t)| \ dt$

$r'(t)=(cos(t), \ -sin(t), \ 2sin(t)cos(t)) \Rightarrow \ |r'(t)|=\sqrt{1+4sin^2(t)cos^2(t)}$ $=\sqrt{1+sin^2(2t)}$

So I have to work out $\int_a^b (2sin(2t), \ sin(2t), \ sin(t)) \sqrt{1+sin^2(2t)} \ dt$

= $\int_a^b (2sin(2t) \sqrt{1+sin^2(2t)}, sin(2t) \sqrt{1+sin^2(2t)}, sin(t) \sqrt{1+sin^2(2t)} \ dt$

From here is it okay to integrate each of the $x,y,z$ values with respect to $t$ ?

Here's what happens when I do (this is for the x coordinate. I thought it would be easier doing each one separately):

Let $u=1+sin^2(2t) \Rightarrow \ \frac{du}{dt}=4sin(2t)cos(2t)$

This gives: $\int_a^b \frac{1}{2}\frac{\sqrt{u}}{cos(2t)}~du$

But we also know that $u=1+(1-cos^2(2t))=2-cos^2(2t) \Rightarrow \ cos(2t)=\sqrt{2-u}$

So the integral for the x coordinate is:

$\frac{1}{2}\int_a^b \sqrt{ \frac{u}{2-u}}~du$ ....which I can't think of a way of integrating

**HallsofIvy** · April 24th 2009, 02:48 AM

Your final integral is clearly wrong: you have an integral of a vector when it should be the integral of a scalar function. You want $\vec{r}\cdot\vec{dr}$ rather than $\vec{r}|\vec{dr}|$ which is what you have.

Also, you need to integrate along the given curve from t= 0 to $t= \pi$ , then integrate along the straight line from (0, -1, 0) back to (0, 1, 0).

**Showcase_22** · April 24th 2009, 03:07 AM

Originally Posted by HallsofIvy

Your final integral is clearly wrong: you have an integral of a vector when it should be the integral of a scalar function. You want $\vec{r}\cdot\vec{dr}$ rather than $\vec{r}|\vec{dr}|$ which is what you have.

Okay, so I need something like:

$\int_C \underline{v} \cdot d \underline{r}=\int_a^b \underline{v} (\underline{r}(t)) \cdot r'(t)~dt$ ?

In which case I would have:

$\int_a^b (2sin(2t), \ sin(2t), \ sin(t)) \cdot (cos(t), \ -sin(t), \ 4sin(2t)cos(2t))~dt$

Also, you need to integrate along the given curve from t= 0 to $t= \pi$ , then integrate along the straight line from (0, -1, 0) back to (0, 1, 0).

So I should have a double integral? ( which makes me question what i've put above this

)

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