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Found this question in my study guide but the answer I got was wrong but I don't see how come.

$$\int^{(1,1,1)}_{(0,0,0)}3x^2y^{10}z^{10}dx+10x^3y^9z^{10}dy+10x^3y^{10}z^9dz$$

I took the integral of each variable giving me:

$$\int^{(1,1,1)}_{(0,0,0)}3x^2y^{10}z^{10}dx+\int^{(1,1,1)}_{(0,0,0)}10x^3y^9z^{10}dy+\int^{(1,1,1)}_{(0,0,0)}10x^3y^{10}z^9dz$$

$$x^3y^{10}z^{10}+x^3y^{10}z^{10}+x^3y^{10}z^{10}=3x^3y^{10}z^{10}$$

$$3x^3y^{10}z^{10}\bigg|^{(1,1,1)}_{(0,0,0)}=3$$ $$Answer=1$$

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    $\begingroup$ There should not be a $3$ in $x^3 y^{10} z^{10}$. $\endgroup$ – user99914 Dec 14 '17 at 2:28
  • $\begingroup$ When you integrate the first part of the trinomial wrt x it will give you $x^3y^{10}z^{10}$, if you continue to integrate the second part of the trinomial wrt y it will give you $x^3y^{10}z^{10}$ as well, continue for the third part and the answer is also $x^3y^{10}z^{10}$. Thus $3x^3y^{10}z^{10}$ $\endgroup$ – Samuel Uribe Dec 14 '17 at 2:29
  • $\begingroup$ Have a look at math.stackexchange.com/questions/1971225/… for why you can’t split a line integral up like that. $\endgroup$ – amd Dec 14 '17 at 8:35
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In general, when evaluating a line integral, you must choose a path along which to integrate. Once your path is chosen, then you can parameterize it and then perform a simple 1D integral.

In this case, we have $$\int_{(0,0,0)}^{(1,1,1)} 3x^2y^{10}z^{10}dx + 10x^3y^9z^{10}dy + 10x^3y^{10}z^9 dz$$

We need to choose a path from the point $(0,0,0)$ to the point $(1,1,1)$ along which to integrate. The most straightforward choice is the straight line connecting them - I can parameterize that path as follows: $$x = t, \ \ dx = dt$$ $$y = t, \ \ dy = dt$$ $$z = t, \ \ dz = dt$$ where $t \in [0,1]$. Using such a parameterization, the integral becomes $$\int_0^1 (3t^{22}+10t^{22}+10t^{22})dt = \int_0^1 23t^{22} dt = t^{23}|^1_0 = 1$$

I didn't have to choose a straight line. I could have chosen a different path and/or parameterization. How about this one?

$$x = t^2, \ \ dx = 2t dt$$ $$y = t^3, \ \ dy = 3t^2 dt$$ $$z = t^5, \ \ dz = 5t^4 dt$$

where $t\in [0,1]$. If I do this, then the integral becomes

$$\int_0^1 3(t^2)^2(t^3)^{10}(t^5)^{10}(2tdt)+10(t^2)^3(t^3)^9(t^5)^{10}(3t^2dt) + 10(t^2)^3(t^3)^{10}(t^5)^9(5t^4dt)$$ $$ = \int_0^1 (6t^{85}+30t^{85}+50t^{85})dt = \int_0^1 86t^{85} = t^{86}|^1_0 = 1$$


The fact that these two paths gave me precisely the same answer is no accident. In general, if the integrand can be written as an exact differential, then the value of the integral is independent of the path chosen. If $$ \int_a^b A(x,y,z)dx + B(x,y,z)dy + C(x,y,z)dz = \int_a^b dF$$ for some function $F(x,y,z)$, then $$\int_a^b dF = F(b)-F(a)$$ In this case, we see that $$F = x^3y^{10}z^{10}$$ fits, and plugging in the end points yields the desired result.


Explicitly, you may still not be sure why your approach does not work. You implicitly chose the path consisting of three straight segments - one from $(0,0,0)$ to $(1,0,0)$, then from $(1,0,0)$ to $(1,1,0)$, and then from $(1,1,0)$ to $(1,1,1)$. The reason your answer is incorrect is because along the first segment, both $y$ and $z$ are equal to zero, so this part vanishes. Similarly, along the second segment, we have that $z$ is equal to zero, so this part vanishes too. The only part of the path which contributes is the third segment, along which both $x$ and $y$ are equal to 1. We then have $$\int_0^1 10(1)(1)z^9 dz = \int_0^1 10z^9 dz = z^{10}|^1_0 = 1$$

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  • $\begingroup$ +1 for explaining why splitting up the integral as the OP did fails. $\endgroup$ – amd Dec 16 '17 at 9:32
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The notation may be throwing you off, but what you're actually integrating is a vector function of the form $$ \boldsymbol{F}(x,y,z) = (3x^2y^{10}z^{10}, 10x^3y^9z^{10}, 10x^3y^{10}z^9) $$

over some path $\boldsymbol{r}(t) = (x(t),y(t),z(t))$ whose endpoints are $(0,0,0)$ and $(1,1,1)$.

Formally, the integral is written as $$ \int\boldsymbol{F}\cdot d\boldsymbol{r} = \int \boldsymbol{F}(x,y,z)\cdot \boldsymbol{r}'(t)dt $$

However, the problem chose to write $\boldsymbol{r}'(t) = (x'(t)dt,y'(t)dt,z'(t)dt) = (dx, dy, dz)$, giving the shorthand form

$$ \int (3x^2y^{10}z^{10}, 10x^3y^9z^{10}, 10x^3y^{10}z^9)\cdot (dx,dy,dz)\ dt \\ \int 3x^2y^{10}z^{10}dx + 10x^3y^9z^{10}dy+10x^3y^{10}z^9 dx $$

with that in mind, observe that $\boldsymbol{F} = \nabla g$, where $g$ is a scalar function $$ g(x,y,z) = x^3y^{10}z^{10} $$

this enables the integral to be path-independent. Therefore we can apply the generalized fundamental theorem of calculus $$ \int_{(0,0,0)}^{(1,1,1)} \boldsymbol{F}\cdot d\boldsymbol{r} = g(1,1,1) - g(0,0,0) = 1 $$


Thoughts: Being able to see through the notation is really important here. While this looks like a deceptively simple problem, there really is more than meets the eye. You're required to know about path integrals of vector functions, and that the vector function you're working with is conservative (path-independence). If the "anti-gradient" function does not exist (it's impossible to write $\boldsymbol{F}$ as $\nabla g$), then there isn't enough information to complete the integral (you'd need to know the path).

The fact that each individual integral evaluates to the same term is also a massive hint, as it points to the existence of $g$

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HINT:

Note that we have

$$\begin{align} 3x^2y^{10}z^{10}\,dx+10x^3y^9z^{10}\,dy+10x^3y^{10}z^9\,dz&=d(x^3y^{10}z^{10})\\\\ &=d(f(x,y,z)) \end{align}$$

where $f(x,y,z)=x^3y^{10}z^{10}$ and

$$df(x,y,z)=\frac{\partial f}{\partial x}\,dx+\frac{\partial f}{\partial y}\,dy+\frac{\partial f}{\partial z}\,dz.$$

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  • $\begingroup$ I understand this: $$\begin{align} 3x^2y^{10}z^{10}\,dx+10x^3y^9z^{10}\,dy+10x^3y^{10}z^9\,dz&=d(x^3y^{10}z^{10}) \end{align}$$ however I don't see how that will help me reach an answer. I see that the original trinomial is the del of the right hand side of the equation aka gradient, but the question is asking me to evaluate the differential which has an integral sign in front of it $\endgroup$ – Samuel Uribe Dec 14 '17 at 2:34
  • $\begingroup$ @SamuelUribe Using the chain rule, we have $df(x,y,z)=\nabla f \cdot (\hat x \,dx+\hat y \,dy+\hat z \,dz)$. And what is the integral $\int df$? $\endgroup$ – Mark Viola Dec 14 '17 at 2:36
  • $\begingroup$ I honestly don't understand why I have to use dot product to evaluate an integral that seems very basic $\endgroup$ – Samuel Uribe Dec 14 '17 at 2:40
  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$ – Samuel Uribe Dec 14 '17 at 2:41
  • $\begingroup$ @SamuelUribe You don't have to use a "dot product." Just recognize that $\int d(f) =f$. Now, evaluate $f$ from the initial point (lower limit) to the final point (upper limit). $\endgroup$ – Mark Viola Dec 14 '17 at 2:59

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