Compute $\int_{\left(0,0,0\right)}^{\left(3,4,5\right)}\frac{xdx+ydy+zdz}{\sqrt{x^{2}+y^{2}+z^{2}}}$

Question

Compute the given line integral:

• $$\int_{\left(0,0,0\right)}^{\left(3,4,5\right)}\frac{xdx+ydy+zdz}{\sqrt{x^{2}+y^{2}+z^{2}}}$$

---

Let $P=\frac{x}{\sqrt{x^{2}+y^{2}+z^{2}}},Q=\frac{y}{\sqrt{x^{2}+y^{2}+z^{2}}}, R=\frac{z}{\sqrt{x^{2}+y^{2}+z^{2}}}$ then $$\color{blue}{\frac{\partial P}{\partial y}}=\frac{-xy}{(x^2+y^2+z^2)^{3/2}},\color{green}{\frac{\partial P}{\partial z}}=\frac{-xz}{(x^2+y^2+z^2)^{3/2}}$$$$\color{blue}{\frac{\partial Q}{\partial x}}=\frac{-xy}{(x^2+y^2+z^2)^{3/2}},\color{red}{\frac{\partial Q}{\partial z}}=\frac{-yz}{(x^2+y^2+z^2)^{3/2}}$$$$\color{green}{\frac{\partial R}{\partial x}}=\frac{-xz}{(x^2+y^2+z^2)^{3/2}},\color{red}{\frac{\partial R}{\partial y}}=\frac{-yz}{(x^2+y^2+z^2)^{3/2}}$$

So the vector field is conservative which means there is $f$ such that $\vec \nabla f=F$, I don't know how to continue.

Answer 1

Consider $$f(x,y,z)=\sqrt{x^2+y^2+z^2}+C$$ then $\vec \nabla f=F$ and $$\int_{\left(0,0,0\right)}^{\left(3,4,5\right)}\frac{xdx+ydy+zdz}{\sqrt{x^{2}+y^{2}+z^{2}}}=f(3,4,5)-f(0,0,0)=5\sqrt{2}.$$

Answer 2

If you have checked the completeness then your expression is definitely of the form $$\frac{\partial U}{\partial x}dx + \frac{\partial U}{\partial y}dy + \frac{\partial U}{\partial z}dz$$ for some (potential) function $U$. Hence $$\begin{cases} \dfrac{\partial U}{\partial x} = \dfrac{x}{\sqrt{x^2+y^2+z^2}} \\ \dfrac{\partial U}{\partial y} = \dfrac{y}{\sqrt{x^2+y^2+z^2}} \\ \dfrac{\partial U}{\partial z} = \dfrac{z}{\sqrt{x^2+y^2+z^2}} \end{cases}$$ Integrating the first equality w.r.t $x$ we have $$\int \dfrac{\partial U}{\partial x}dx = \int\dfrac{x}{\sqrt{x^2+y^2+z^2}}dx$$ substitute $t(x) = \sqrt{x^2+y^2+z^2}$ in the integral on the RHS. Then, $$dt = \dfrac{x}{\sqrt{x^2+y^2+z^2}}dx$$ and therefore $$\begin{align} U &= \int dt \\ &= t + C(y,z) \\ &= \sqrt{x^2+y^2+z^2} + C(y,z) \tag{1}\end{align}$$ Similarly, $$U = \sqrt{x^2+y^2+z^2} + C(x,y) \tag{2}$$ $$U = \sqrt{x^2+y^2+z^2} + C(z,x) \tag{3}$$ Now, you show that $C(x,y) = \text{const}$.

Answer 3

Compute the given line integral:

$$\int_{\left(0,0,0\right)}^{\left(3,4,5\right)}\frac{xdx+ydy+zdz}{\sqrt{x^{2}+y^{2}+z^{2}}}$$

This can be made a lot simpler by change of coordinates into spherical. Recall that:

$$ \rho= \sqrt{x^2 +y^2 +z^2}$$

This leads to:

$$ d \rho = \frac{x dx + y dy + zdz}{\sqrt{x^2 + y^2 +z^2}}$$

Hence, our integral just becomes to evaluate:

$$ \rho(3,4,5) - \rho(0,0,0)=5\sqrt{2}$$