Step up to the plate

It is commonly known that the mass of an object is defined as:

$\text{Mass = Density} * \text{Volume}$

or in Calculus terms:

$\Large{M = \iiint _{ Q }^{ }{ \rho (x,y,z) } \, dV }$

$\rho (x,y,z) \text{: density function}$

$dV \text{: differential volume}$

$Q \text{: bounded volume}$

However, since we are dealing in the 2 dimensional xy plane the equation reduces to:

$\Large{M = \iint _{ R }^{ }{ \rho (x,y) } \, dA = \int _{ a }^{ b }{ \int _{ { h }_{ 1 }(x) }^{ { h }_{ 2 }(x) }{ \rho (x,y) \, dydx } } }$

where the bounds of x,y are are:

$\large{a\leq x\leq b}$ and $\large{h_{1}(x) \leq y \leq h_{2}(x)}$

The x bounds of this problem can simply be found as the x-coordinates of the intersection between the functions (hence, their equality):

$\large{x = x^{2} \Rightarrow x = 0,1}$

The y bounds of the inner integral are represented as a function of x, and as usual, you must always subtract the lower function from the upper one to find the positive area.

And in this case:

$\large{{ x }^{ 2 } \le x \quad\quad \{ x\mid 0\le x\le 1\} }$

Given our density function:

$\large{ {\rho (x,y)}= 1 +xy}$

Substituting into our earlier double integral equation we get:

$\Large{\int _{ 0 }^{ 1 }{ \int _{ {x}^{2} }^{ x }{ 1 +xy \,\, dydx } } }$

Integrating with respect to y (remember to treat x as a constant) :

$\Large {\Rightarrow\int _{ 0 }^{ 1 }{ \left. (y+\dfrac { { xy }^{ 2 } }{ 2 } )\right|_{x^{2}}^x dx } }$

Substituting in the bounds of the inner integral:

$\Large {\Rightarrow\int _{ 0 }^{ 1 }{ ((x) + \frac{ {x(x)}^{2} } {2} } ) - ((x^{2}) + \frac{ {x(x^{2})}^{2} }{2} ) \,dx}$

And after some simplification:

$\Large { \Rightarrow \int _{ 0 }^{ 1 }{ \frac { {-x}^{5} }{ 2 } +\frac { {x}^{3} }{ 2 } -{x}^{2} +x \, dx } }$

Further integrating once more, but this time with respect to x:

$\Large{ \Rightarrow\left.( \frac { {-x}^{6} }{ 12 } +\frac { {x}^{4} }{ 8 } -\frac { {x}^{3} }{ 3 } +\frac { {x}^{2} }{ 2 } )\right|_0^1}$

Substituting the final bounds and approximating:

$= \large{ \frac{1}{2} +\frac{1}{8} - \frac{1}{3} - \frac{1}{12} = \frac{5}{24} \approx \boxed{0.20833333..} }$

This is one of my few solutions, so please let me know of any improvements I can make =)

$Intersection~Points, ~y=x ~and~ y=x^2,\therefore~x=x^2,\\\implies x=0, ~x=1.~~\therefore~the~two~~intersection~points~are~(0,0),(1,1).$

$dM=(1+xy)dA=(1+xy)\delta x*\delta y. \\ \text {In the vertical strip } \delta x~\text{ moves down from } y=x,~ to~y=x^2,~~\\ \\\therefore \text{ the strip itself moves from x=0 to x=1......}~~~\\\implies~\Large \int _ 0^1 \int _{y=x^2 }^{y= x } \{ 1 +xy \} dydx =\dfrac 5 {25}=\color{#D61F06}{0.20833} \\\text{For dxdy, in the horizontal strip } \delta y ~moves~ from~ x=y~ to ~x~ = \sqrt y.\\\text{and the strip from 0 to 1, since intersect is at (1,1)}.\\\implies~\Large \int _ 0^1 \int _{x=\sqrt y }^{x= y } \{ 1 +xy \} dydx =\dfrac 5 {25}=\color{#D61F06}{0.20833}$

The stuff you can do with random sampling is pretty neato.

Python 2.7:

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from random import random
from time import time
def in_bounds(x, y):
    return y <= x and y >= x*x
def D(x, y):
    return 1 + x*y
total_area = 0.0
total_density = 0.0
area_trials = 0.0
density_trials = 0.0
end = time() + 10
while time() < end:
    # assign x and y a random real
    # between 0 and 1
    x, y = random(), random()
    if in_bounds(x, y):
        total_area += 1
        total_density += D(x, y)
        density_trials += 1
    area_trials += 1
area = total_area / area_trials
density = total_density / density_trials
print "Mass of plate:", area * density

Also, good to see you're still writing problems!