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

#### Prove It

##### Well-known member
MHB Math Helper
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

##### Well-known member
MHB Math Helper
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.)

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#### HallsofIvy

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

In that case, we must have $$\displaystyle \frac{\partial F}{\partial x}= 6x^2$$ so that $$\displaystyle F= 2x^3$$ 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, $$\displaystyle F(x,y,z)= 2x^3+ G(y, z)$$.

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

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

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

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

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

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