MHB Luca's question via email about a line integral....

[Prove It]

Evaluate the line integral $\displaystyle \begin{align*} I = \int_{(0,0,0)}^{\left( 5, \frac{1}{2}, \frac{\pi}{2} \right) }{ 6\,x^2\,\mathrm{d}x + \left[ 6\,z^2 + 9\,\mathrm{e}^{9\,y} \cos{ \left( 10\,z \right) } \right] \,\mathrm{d}y + \left[ 12\,y\,z - 10 \,\mathrm{e}^{9\,y}\sin{ \left( 10\,z \right) } \right] \,\mathrm{d}z } \end{align*}$

I am assuming that this line integral is along the straight line from $\displaystyle \begin{align*} (0,0,0) \end{align*}$ to $\displaystyle \begin{align*} \left( 5, \frac{1}{2}, \frac{\pi}{2} \right) \end{align*}$, which has equation $\displaystyle \begin{align*} \left( x, y, z \right) = t\left( 5, \frac{1}{2} , \frac{\pi}{2} \right) \, , \,t \in \left[ 0, 1 \right] \end{align*}$, so

$\displaystyle \begin{align*} x &= 5\,t \implies \mathrm{d}x = 5\,\mathrm{d}t \\ y &= \frac{1}{2}\,t \implies \mathrm{d}y = \frac{1}{2}\,\mathrm{d}t \\ z &= \frac{\pi}{2}\,t \implies \mathrm{d}z = \frac{\pi}{2}\,\mathrm{d}t \end{align*}$

and so the integral becomes

$\displaystyle \begin{align*} I &= \int_{t=0}^{t=1}{ 6\left( 5\,t \right) ^2 \cdot 5\,\mathrm{d}t + \left[ 6\left( \frac{\pi}{2}\,t \right) ^2 + 9\,\mathrm{e}^{9 \cdot \frac{1}{2}\,t} \cos{ \left( 10 \cdot \frac{\pi}{2}\,t \right) } \right] \cdot \frac{1}{2}\,\mathrm{d}t + \left[ 12 \cdot \frac{1}{2}\,t \cdot \frac{\pi}{2}\,t - 10 \,\mathrm{e}^{ 9 \cdot \frac{1}{2}\,t } \sin{ \left( 10\cdot \frac{\pi}{2}\,t \right) } \right] \cdot \frac{\pi}{2}\,\mathrm{d}t } \\ &= \int_0^1{ \left[ 750\,t^2 + \frac{3\,\pi ^2}{4}\,t^2 + \frac{9}{2}\,\mathrm{e}^{\frac{9}{2}\,t} \cos{ \left( 5\,\pi\,t \right) } + \frac{3\,\pi ^2}{2} \, t^2 - 5\,\pi\,\mathrm{e}^{\frac{9}{2}\,t}\sin{\left( 5\,\pi\,t \right) } \right] \,\mathrm{d}t } \\ &= \int_0^1{ \left[ \left( \frac{3000 + 9\,\pi^2}{4} \right) t^2 + \frac{9}{2}\,\mathrm{e}^{\frac{9}{2}\,t}\cos{\left( 5\,\pi\,t \right) } - 5\,\pi\,\mathrm{e}^{\frac{9}{2}\,t}\sin{ \left( 5\,\pi\,t \right) } \right] \,\mathrm{d}t } \\ &= \left\{ \left( \frac{1000 + 3\,\pi^2}{4} \right) t^3 + \frac{9}{2} \left[ \frac{\mathrm{e}^{\frac{9}{2}\,t}}{\left( \frac{9}{2} \right) ^2 + \left( 5\,\pi \right) ^2} \right] \left[ \frac{9}{2} \, \cos{ \left( 5\,\pi\,t \right) } + 5\,\pi \sin{ \left( 5\,\pi\,t \right) } \right] - 5\,\pi \left[ \frac{ \mathrm{e}^{ \frac{9}{2} \,t } }{ \left( \frac{9}{2} \right) ^2 + \left( 5\,\pi \right) ^2 } \right] \left[ \frac{9}{2} \, \sin{ \left( 5\,\pi\,t \right) } - 5\,\pi \cos{ \left( 5 \, \pi \, t \right) } \right] \right\} _0^1 \\ &= \left\{ \left( \frac{1000 + 3\,\pi ^2}{4} \right) t^3 + \left[ \frac{4\,\mathrm{e}^{\frac{9}{2}\,t}}{81 + 100\,\pi^2} \right] \left[ \left( \frac{81 + 100\,\pi ^2}{4} \right) \cos{ \left( 5\,\pi\,t \right) } \right] \right\} _0^1 \\ &= \left[ \left( \frac{1000 + 3\,\pi ^2}{4} \right) t^3 + \mathrm{e}^{\frac{9}{2}\,t} \cos{ \left( 5\,\pi\,t \right) } \right] _0^1 \\ &= \frac{1000 + 3\,\pi ^2}{4} - \mathrm{e}^{\frac{9}{2}} - 1 \\ &= \frac{996 + 3\,\pi ^2 - 4\,\mathrm{e}^{\frac{9}{2}}}{4} \\ &\approx 166.385\,072 \end{align*}$

 

[HallsofIvy]

Assuming that the integral is independent of the path (which you would need to do since no path is specified) another good choice would be the line, along the x-axis, from (0, 0, 0) to (5, 0, 0) then the line, parallel to the y-axis, from (5, 0, 0) to (5, 1/2, 0), then along the line, parallel to the z-axis, from (5, 1/2, 0) to (5, 1/2, pi/2).

On the first part, only x changes so dy and dz are 0. The integral becomes
\int_0^5 6x^2 dx= \left[2x^3\right]_0^5= 250.

On the second part, only y changes so dx and dz are 0. x is the constant, 5, and z is 0. The integral becomes \int_0^{1/2} 9e^{9y}dy= \left[e^{9y}\right]_0^{1/2}= e^{9/2}- 1.

On the third part, only z changes so dx and dy are 0. x is the constant, 5, and y is the constant, 1/2. The integral becomes \int_0^{\pi/2} 6z+ 10e^{9/2} sin(10z)dz= \left[3z^2- e^{9/2}cos(10z)\right]_0^{\pi/2}= \frac{3\pi^2}{4}+ 2e^{9/2}.

The original integral is the sum of those, 249+ \frac{3\pi^2}{4}+ 3e^{9/2}.

(Better check my arithmetic.)

 

[HallsofIvy]

[FONT=MathJax_Size1]Of course, if the integral is independent of the path then the integrand, [math]6x^2 dx+ (6z^2+ 9e^{9y}cos(10z))dy+ (12yz- 10e^{9y} sin(10z)dz[/math], must be "exact"- that is, there exist some differentiable function, F(x, y, z), such that the differential is [math]dF= 6x^2 dx+ (6z^2+ 9e^{9y}cos(10z))dy+ (12yz- 10e^{9y} sin(10z)dz[/math] and the integral is just F evaluated at the limits of integration.

In that case, we must have [math]\frac{\partial F}{\partial x}= 6x^2[/math] so that [math]F= 2x^3[/math] plus a "constant". But since the differentiation is with respect to x only, that "constant" can be an arbitrary function of y and z. That is, [math]F(x,y,z)= 2x^3+ G(y, z)[/math].

Differentiating with respect to y, [math]\frac{\partial F}{\partial y}= \frac{\partial G}{\partial y}[/math] and that must be equal to [math]6z^2+ 9e^{9y}cos(10z)[/math].

Integrating [math]\frac{\partial G}{\partial y}= 6z^2+ 9e^{9y}cos(10z)[/math] with respect to y, [math]G(y,z)= 6yz^2+ e^{9y}cos(10z)+ H(z)[/math] where, now, the "constant of integration" must be a differentiable function of z only.

So [math]F(x,y,z)= 2x^3+ 6yz^2+ e^{9y}cos(10z)+ H(z)[/math]. Differentiating that with respect to z, [math]\frac{\partial F}{\partial z}= 12yz- 10e^{9y}sin(10z)+ \frac{dH}{dz}= 12yz- 10e^{9y} sin(10z)[/math] so that \frac{dH}{dz}= 0 and H really is a constant. (And this integral really is independent of the path.)

We have [math]F(x,y,z)= 2x^3+ 6yz^2+ e^{9y}cos(10z)+ C[/math] and need to evaluate that at [math](0, 0, 0)[/math] and [math](5, 1/2, \pi/2)[/math].

[math]F(0,0,0)= 1+ C[/math] and [math]F(5,1/2,\pi/2)= 250+3\pi^2/4- e^{9/2}+ C[/math] so the integral is [math]249+ \frac{3\pi^2}{5}- e^{9/2}[/math]