integral of absolute value of a function. The problem is as following:
straight lines are drawn joining every point on $y=\sqrt{3}x\sin{x}$ to the origin. what fraction of the x-y plane is occupied by these lines.
My solution is:
I observe that the limit of $y=\sqrt{3}x\sin{x}$ is oscillating between positive and negative. So the graph of the function cross x-axis infinitely many times as $x$ goes to infinity. What the problem ask is actually to compute the fraction of the integral of absolute value of the funciton to the area of the $xy$-plane. Since $y=\sqrt{3}x\sin{x}$ is even function, it is enough to consider right half of $xy$-plane. I compute the integral of absolute value of $y=\sqrt{3}x\sin{x}$ on the right half plane by spliting the right half $xy$-plane by $k\pi$'s, where $k$ represents an integer range from $0$ to infinity. 
$\frac{\sum_{k=0}^{\infty}|\int_{k\pi}^{(k+1)\pi}\sqrt{3}x\sin{x} dx|}{2\sum_{k=0}^{\infty}\pi\lim_{x\rightarrow \infty}{x}}$
After computation, the result on the top of the fraction is $\sqrt{3}\sum_{k=0}^{\infty}(2k+1)\pi$ and the result on the bottom of the fraction is $\lim_{n\rightarrow\infty}(2n\pi\lim_{x\rightarrow \infty}{x})$. 
Hence the fraction is result of the fraction is $\lim_{n\rightarrow\infty}\frac{\sqrt{3}(n+1)^2}{2n{\lim_{x\rightarrow \infty}{x}}}$, which is just to compute the limit of the expression $\lim_{x\rightarrow \infty}\frac{\sqrt{3}(x+1)^2}{2x^2}=\frac{\sqrt{3}}{2}$.
But the place I find this question, it is with an answer that is $\frac{2}{3}$.
So I am wondering if these are any mistakes of my solution? or I am correct and the answer I found is wrong?
 A: I think you complicated your life.

Here is the graph of $\sqrt{3}x\sin(x)$.
Since $|\sin(x)|<1$ then $-\sqrt{3}x\le f(x)\le\sqrt{3}x$ and the set of lines joining every point of the curve to the origin is just the part of the plane between these two straight lines (imagine the blue part, all filled with blue).
Since this region is infinite as well as the plane, I guess the ratio is calculated by evaluating the angular filling.
The angle of $y=\pm\sqrt{3}x$ are $\pm 60°$ so the overall angular load factor is $\dfrac{4\times60°}{360°}=\dfrac 23$
A: Checking your answer without using calculus:
Since $\sin x$ varies continuously from its minimum, $-1$, to its maximum, $1$, the drawn rays have slopes from $-\sqrt{3}$ to $\sqrt{3}$.  These rays are drawn both to the left and the right of the origin.  The ray with slope $\sqrt{3}$ to the right of the origin makes an angle of $\pi/3$ with the positive $x$-axis.  The ray with slope $-\sqrt{3}$ likewise makes an angle of $\pi / 3$.  To the left of the origin, the same is obtained.  This means the region(s) covered by rays fill $4 \cdot \frac{\pi}{3}$.  Thus, they cover $\frac{4 \pi /3}{2 \pi} = \frac{4}{6} = \frac{2}{3}$ of the plane.
Your solution has a few defects.  


*

*You do not include the region to the left of the origin.

*When you integrate $\sqrt{3} x \sin x$, you are finding the area covered by lines from the point on the graph extending vertically to the $x$-axis, not from the point on the graph to the origin.

*Your denominator is nonsense.  It says: first find $\lim_{x \rightarrow \infty} x = \infty$, then multiply that $\pi$, yielding $\infty$, then sum that for $k=0$ to $\infty$, obtaining $\infty$, finally, multiply by $2$, producing $\infty$.

*If you were to use this method, your denominator should be a rectangle that grows in both $x$- and $y$-directions with $k$.


If we were going to attack the question you answered (which is not the given question), we would have tried something like
$$  \lim_{M \rightarrow \infty} \frac{\int_{-M}^M \; | \sqrt{3} x \sin x |  \,\mathrm{d}x}{(2M)(2 \sqrt{3} M)}  \text{.}  $$
This version


*

*Includes the part of the plane to the left of the origin.

*Integrates the length of each line segment, rather than trying to split the range of integration into pieces where the line segments are all either positively directed or negatively directed.

*Finds the part of the plane covered over the range $x \in [-M,M]$, divided by the part of the plane $[-M,M] \times [-\sqrt{3} M, \sqrt{3} M]$, the rectangle bounding the envelope of the function.

*Puts the limit in a place where it can work for us, and shows that we are looking at the sequence of fractions of bounding rectangles covered by vertical lines from the graph of the function to the $x$-axis.


That integral is a little tricky.  If we split into regions bounded by integer multiples of $\pi$, we get 
$$  \int_{k \pi}^{(k+1)\pi} \; \sqrt{3} x \sin x \, \mathrm{d} x = (-1)^k \sqrt{3}\pi|2k+1|  \text{.}  $$
When $k$ is even, these contributions are positive, and when $k$ is odd, negative, so we can correct the signs (to implement the absolute value) by dropping the "$(-1)^k$".  Since we are now breaking the $x$-axis up into $pi$ wide pieces, we should adjust the terms in the denominator to do so as well.  Making these replacements, we obtain \begin{align*}
  \lim_{\substack{M \rightarrow \infty\\M \in \mathbb{Z}}} & \frac{\sum_{k=-M}^{M} \sqrt{3} \pi |2k+1|}{(2 \pi M)(2 \sqrt{3} \pi M)}  \\
    &= \lim_{\substack{M \rightarrow \infty\\M \in \mathbb{Z}}} \frac{\sqrt{3}\pi(2M^2 + 2M + 1)}{(2 \pi M)(2 \sqrt{3} \pi M)}  \\
    &= \frac{1}{2\pi}  \text{.}
\end{align*}
This is substantially smaller than $\frac{2}{3}$ because only counting the area between the graph of the function and the $x$-axis captures substantially less area than counting the area covered by rays from the origin to the points of the graph.
A: Let me explain myself in another way by answering a question: What fraction of $xy$-plane occupied by the area bounded by $|\sqrt{3}x\sin{x}|$ and the $x$-axis?
After computing the integral of $|\sqrt{3}x\sin{x}|$ as what I did in my first post, the solution leads to compute the limit of 
$\frac{\sqrt{3}\lim_{M\rightarrow\infty}(M+1)^2}{2\lim_{N\rightarrow \infty}N^2}$
 where both top and bottom are discrete functions. 
For this question, I think we can choose any function, to compute the area of the $xy$-plane ( as long as this function is able to get the area. e.g we can choose either discrete or continuous function). But for this specific problem, we'd better to use discrete function, because the top of the fraction is discrete. In order to use the quotient rule for limit, we require that both top and bottom of the quotient to have the same property (both discrete or both continuous). 
How can we use the quotient rule to calculate limit of a function with multiple variables? This arises to same problem as to compute the limit:
$\lim_{x\rightarrow 0, y\rightarrow 0}\frac{xy}{x^2+y^2}$
We can solve this question by change the Cartesian coordinate to polar coordinate. But for the question here, this technique is not applicable. 
I compute the limit of this quotient with different relation between $M$ and $N$. For example use linear relation $N=\sqrt{3}M$ or $N=M^2$. We get different answer. 
Conclusion: The limit of this quotient do not exist. the theorem I use is that a function can only have one limit if its limit exists. 
Hence, the solution of this question does not exist.
We should not change different variables directly to the same variables. As this changes, we narrow our computation to a fix relation between two variables. Actually, if not specifically mentioned, there will be no relation between variables. in this question, there is no extra conditions mentioned to restrict the variable for the functions on the top of the quotient and on the bottom of the quotient to only one relation: $M=\sqrt{3}N$. We should not add a relation ourselves, which may cause incorrect judgement for the existence of the limit.
