I had previously asked a similar version of this question, but realized I phrased it incorrectly as there was a divergence issue with the integral, but now that that's been fixed:
Find the expected magnitude of the sum of a unit vector $U$ and a vector $V$ of magnitude $\sqrt3$ in the Cartesian plane, given that the $x-$component of $U$ is randomly chosen from the interval [−1,1], and that the $x-$component of $V$ is randomly chosen from the interval [−sqrt3,sqrt3].
A.) $\dfrac{28\sqrt3}{27}$ B.) $\dfrac{10\sqrt3}9$ C.) $\dfrac{4\sqrt3}3$ D.) $2$
I started by writing out an integral that expressed the expected value of the magnitude. However, I was unsure what to use for f(x) and do not know how to proceed, I ended up with something like the integral from negative to infinity of xf(x)dx. I should note that this is a problem I came up on my own with and want to see the different ways of doing it. One of my friends solved it using Multivariable Calculus (set up a double integral), but I do believe that was overkill. Nevertheless, any help or sketches of solutions would be appreciated!
This is the double integral I get: 
For the problem as stated, none of the choices work.
In other words, the problem is wrong.
To get an understanding of what went wrong, I'll change the problem so as to
Afterwards, I'll deal with the actual problem.
For the simpler problem, fix $U=\langle{1,0}\rangle$.
Let $V=\langle{x,y}\rangle$.
From the given information, we have $$ \begin{cases} -\sqrt{3} \le x \le \sqrt{3} \qquad\qquad\qquad\qquad\qquad\;\;\;\;\;\, \\[4pt] y^2=3-x^2 \end{cases} $$ hence \begin{align*} |U+V| &= |\langle{1+x,y}\rangle| \qquad\qquad\qquad\qquad\qquad\; \\[4pt] &= \sqrt{(1+x)^2 + y^2}\\[4pt] &= \sqrt{ (1+2x + x^2)+(3-x^2)}\\[4pt] &= \sqrt{4+2x}\\[20pt] \implies\;E(|U+V|) &= \frac{1}{2\sqrt{3}}\int_{-\sqrt{3}}^{\sqrt{3}}\,\sqrt{4+2x}\,dx = \frac{10\sqrt{3}}{9} \approx 1.924500898 \end{align*} Now for the actual stated problem . . .
Suppose the unit vector $U=(a,b)$, where $a$ is chosen uniformly at random in the interval $[-1,1]$.
By symmetry, we can assume $b \ge 0$, but we can't assume $U = \langle{1,0}\rangle$.
The expected value of $|U+V|$ is not independent of the choice of $a$.
Before doing the full analysis, let's take one more example, to verify the lack of independence.
Thus, take $U = \langle{0,1}\rangle$.
Consider two cases . . .
Case $(1)$:$\;y \ge 0$. \begin{align*} \text{Then}\;\;|U+V| &= |\langle{x,1+y}\rangle| \qquad\qquad\qquad\qquad\qquad\; \\[4pt] &=\sqrt{x^2 + (1+y)^2}\\[4pt] &=\sqrt{x^2 + (1 + \sqrt{3-x^2})^2}\\[4pt] &=\sqrt{2x^2+1+ 2\sqrt{3-x^2}}\\[20pt] \implies\;E(|U+V|) &= \frac{1}{2\sqrt{3}}\int_{-\sqrt{3}}^{\sqrt{3}}\,\sqrt{2x^2+1+ 2\sqrt{3-x^2}}\,dx \approx 2.587781334 \end{align*} Case $(2)$:$\;y \le 0$. \begin{align*} \text{Then}\;\;|U+V| &= |\langle{x,1+y}\rangle| \qquad\qquad\qquad\qquad\qquad\; \\[4pt] &=\sqrt{x^2 + (1+y)^2}\\[4pt] &=\sqrt{x^2 + (1 - \sqrt{3-x^2})^2}\\[4pt] &=\sqrt{2x^2+1- 2\sqrt{3-x^2}}\\[20pt] \implies\;E(|U+V|) &= \frac{1}{2\sqrt{3}}\int_{-\sqrt{3}}^{\sqrt{3}}\,\sqrt{2x^2+1- 2\sqrt{3-x^2}}\,dx \approx 1.085496494 \end{align*} Averaging the results for the two cases, we get $$E(|U+V|) \approx 1.836638914$$ which is not the same as the result obtained using $U=\langle{1,0}\rangle$.
Next, let $a$ vary uniformly in the interval $[0,1]$, so as to get the true (approximate) answer for the actual stated problem .
Without loss of generality, we can assume $b \ge 0$.
Consider two cases . . .
Case $(1)$:$\;y \ge 0$. \begin{align*} \text{Then}\;\;|U+V| &= |\langle{a+x,b+y}\rangle| \qquad\qquad\qquad\qquad\qquad\; \\[4pt] &=\sqrt{(a+x)^2+ (b+y)^2}\\[4pt] &=\sqrt{(a+x)^2+ (\sqrt{1-a^2} + \sqrt{3-x^2})^2}\\[20pt] \implies\;E(|U+V|) &= \frac{1}{4\sqrt{3}} \int_{-\sqrt{3}}^{\sqrt{3}}\, \int_{-1}^{1} \sqrt{(a+x)^2+ (\sqrt{1-a^2} + \sqrt{3-x^2})^2} \,da\,dx \approx 2.454948895 \end{align*} Case $(2)$:$\;y \le 0$. \begin{align*} \text{Then}\;\;|U+V| &= |\langle{a+x,b+y}\rangle| \qquad\qquad\qquad\qquad\qquad\; \\[4pt] &=\sqrt{(a+x)^2+ (b+y)^2}\\[4pt] &=\sqrt{(a+x)^2+ (\sqrt{1-a^2} - \sqrt{3-x^2})^2}\\[20pt] \implies\;E(|U+V|) &= \frac{1}{4\sqrt{3}} \int_{-\sqrt{3}}^{\sqrt{3}}\, \int_{-1}^{1} \sqrt{(a+x)^2+ (\sqrt{1-a^2} - \sqrt{3-x^2})^2} \,da\,dx \approx 1.275057141 \end{align*} Averaging the results for the two cases, we get $$E(|U+V|) \approx 1.865003018$$ The above result is the true (approximate) answer to the actual stated problem, but it doesn't match any of the choices.