Evaluate $\int_c \mathbf F\bullet d\mathbf r$ along a given path

Question

I've been stuck on one of our practise problems for Engineering Mathematics all day and am unsure how to tackle this problem.

The problem asks to evaluate $\int_C \mathbf F\cdot d\mathbf r$ along a given path, considering

$$\mathbf F(x,y,z)=(e^x\sin y)\mathbf i+(e^x\cos y)\mathbf j+z^2\mathbf k$$ and $$C=(\sqrt{t},t^3,e^\sqrt{t})$$ for $0\le t\le 1$.

Now I understand that to solve this problem we need to integrate the following:

$\int_0^1 \mathbf F(x,y,z)\cdot {d\mathbf r\over dt}\,dt$

Where $r(t)\equiv\mathbf c=(\sqrt{t},t^3,e^\sqrt{t})$.

Hence the integral should look like:

$\int_0^1 [(e^\sqrt{t}\sin (t^3)\cdot {1\over {2\sqrt{t}}})\,+\,(e^\sqrt{t}\cos (t^3)\cdot 3t^2)\,+\,(({e^\sqrt{t}})^2\cdot {e^\sqrt{t}\over 2\sqrt{t}}\,)]dt$

Which equates to

$\int_0^1 {{e^\sqrt{t}}\sin (t^3)\over{{2\sqrt{t}}}}dt$ + $\int_0^1 3t^2e^\sqrt{t}\cos (t^3)dt$ + $\int_0^1 {({e^\sqrt{t}})^3 \over 2\sqrt{t}}dt$

I don't know if this is the correct integral, but if it is, I'm at a loss as to how to work it out. Would anyone be able to point out if I've interpreted the question correctly and furthermore would anyone be able to point me in the right direction as to which method to use?

I've tried entering this into MATLAB and Mathematica (not as familiar with Mathematica as MATLAB) and it doesn't appear to give me the correct result in both programs, which suggest I haven't entered/calculated the integral properly or I've misinterpreted the entire question.

The final result should be:

$$e\sin(1)+{1\over3}e^3-{1\over3}$$

Any help is appreciated. Cheers!

Answer 1

Note $\mathcal{curl\,}F=0$ because $$\left| \begin{matrix} i & j & k \\ \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\ {{e}^{x}}\sin y & {{e}^{x}}\cos y & {{z}^{2}} \\ \end{matrix} \right|=(0\,\,,0\,\,,\,\,{{e}^{x}}\cos y-{{e}^{x}}\cos y)=(0,0,0)$$ i.e $F$ is Conservative, therefore $$\int_C \mathbf F\bullet d\mathbf r=\Phi(B)-\Phi(A)$$ now we should find $\Phi$ $$\Phi(x,y,z)=\int\,e{^x}sin\,y\,dx+0+\int z^2dz=e{^x}sin\,y+\frac{1}{3}z^3$$ then $$\int_C \mathbf F\bullet d\mathbf r=\Phi(1,1,e)-\Phi(0,0,1)=e\,sin(1)+\frac{1}{3}e^3-\frac{1}{3}$$ Note $$A=C(\sqrt{0},0^3,e^{\sqrt{0}})=(0,0,1)$$ $$B=C(\sqrt{1},1^3,e^{\sqrt{1}})=(1,1,e)$$

Answer 2

The formal definition of the path integral is:

$$\int_c \vec{F}\cdot d\vec{r}=\int_a^b\vec{F}(\gamma(t)) \cdot \dot{\gamma(t)} dt$$

It basically translates to: You have some parameterization $\gamma(t)$, plug that parameterization into your vector field, calculate the dot product and then evaluate the integral:

In your case the parameterization is: $$\gamma(t)=\begin{cases}x=\sqrt{t} \\y=t^3 \\ z=\sqrt{t}\end{cases} \space \space \space \space \space 0\le t \le 1 $$

and

$$\dot{\gamma}(t)=\begin{cases}x=\frac{1}{2 \sqrt{t}} \\y=3t^2 \\ z=\frac{1}{2 \sqrt{t}}\end{cases} \space \space \space \space \space 0\le t \le 1 $$

Plugged into your vector field you get:(I believe it is here where you made a mistake)

$$\vec{F}({\gamma(t)})=\begin{pmatrix} e^{\sqrt{t}} \sin{t^3} \\e^{\sqrt{t}} \cos{t^3} \\e^{2\sqrt{t}}\end{pmatrix}$$

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The integral becomes:

$$\int_0^1 \begin{pmatrix} e^{\sqrt{t}} \sin{t^3} \\e^{\sqrt{t}} \cos{t^3} \\t\end{pmatrix} \begin{pmatrix}\frac{1}{2 \sqrt{t}} \\ 3t^2 \\ \frac{1}{2 \sqrt{t}} \end{pmatrix}dt=\int_0^1\frac{e^{\sqrt{t}} \sin{t^3}}{2\sqrt{t}}dt+\int_0^1 (e^{\sqrt{t}} \cos{t^3})(3t^2)dt+\int_0^1\frac{e^{2\sqrt{t}}}{2\sqrt{t}}dt$$

I am terrible at integrating (maybe some other users can help you do the integral by hand) but mathematica evaluates the indefinite integral to:(LINK)

$$\int \frac{e^{\sqrt{t}} \sin{t^3}}{2\sqrt{t}}+(e^{\sqrt{t}} \cos{t^3})(3t^2)+\frac{e^{2\sqrt{t}}}{2\sqrt{t}}dt=\frac{1}{2}e^{\sqrt{t}}\left(2 \sin(t^3)+e^{\sqrt{t}}\right)+c$$

However, note that this does not equal to the answer you have provided in your post. Instead, evaluating the integral from $0$ to $1$ leads to:

$$\boxed{e \sin(1)+\frac{e^2}{2}-\frac{1}{2}}$$

So, either your answer is wrong or your vector field or parameterization need to be different.

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Edit:

Assuming the vector field is $z^3$ in the $\mathbf{k}$ component, the integral evaluates to:

$$\int \frac{e^{\sqrt{t}} \sin{t^3}}{2\sqrt{t}}+(e^{\sqrt{t}} \cos{t^3})(3t^2)+\frac{e^{3\sqrt{t}}}{2\sqrt{t}}dt=e^{\sqrt{t}} \sin(t^3)+\frac{e^{3\sqrt{t}}}{3}+c$$

Integrating from $0$ to $1$ gives the desired result:

$$\boxed{e \sin(1)+\frac{e^3}{3}-\frac{1}{3}}$$