Electric charge is distributed
over the triangular region D shown below
so that the charge density at the point x comma y
is given by sigma of x comma y equals two x y,
measured in coulombs per square meter.
We're asked to find the total charge on the region D.
If we look at our notes below,
suppose an electrical charge is distributed over a region R
in the xy-plane and the charge density at a point x comma y
is given by sigma of x comma y,
then the total charge over the region R
is given by this double integral.
So in our case, the total charge Q is equal to
the double integral over the region D,
this triangular region,
and sigma of x comma y is equal to two x y
and we have differential A.
Now let's focus on D, the region of integration.
Notice how the three corner points of this triangular region
would be the points two comma zero,
two comma four,
and zero comma four
and, therefore, the line containing
this side of the triangle
would be x equals two.
This side would be on the line y equals four.
Now for the line containing this side,
notice how the vertical intercept would be four,
and if we go down four and right two,
we'd be at this point in the line
and, therefore, the slope of this line is negative two.
So the equation of the line containing this side
would be y equals either negative two x plus four
or four minus two x.
Differential A is equal to either d x d y or d y d x,
which determines our order of integration.
Let's use d y d x, so we're going to get respect to y first,
then with respect to x.
With respect to y, we want to see which equations
bound this region above and below.
Notice how this region is bounded below
by y equals four minus two x,
so four minus two x is the lower limit of integration
with respect to y.
It's bounded above by y equals four,
so four is the upper limit of integration for y.
Now that we have the region bounded above and below,
we need with respect to x, looking back at our region,
notice how we start at x equals zero
and stop at x equals two.
So the limits of integration with respect to x
are from zero to two.
Let's go and evaluate this on the next slide.
So we first integrate with respect to y,
treating x as a constant,
so the antiderivative is going to be
two x times y squared divided by two.
So, again, we have two x times y squared divided by two.
We can simplify the antiderivative.
Two divided by two simplifies to one,
and with these limits of integration for y,
so we'll substitute these values for y
and then find the difference.
So when y is equal to four,
we would have x times four squared
and then when y is equal to four minus two x,
we'll have x times the quantity four minus two x squared.
So here we'll just have 16 x
and here we're going to have minus x times
the quantity four minus two x squared
is going to be 16 minus 16 x plus 4 x squared.
And now simplifying inside the parentheses,
we have 16 x and then we're going to have minus 16 x
and then we'll have plus 16 x squared
and then we'll have minus four x cubed.
And notice how here, 16 x minus 16 x would be zero.
So now we have the integral from zero to two
of just 16 x squared minus 4 x to the third.
And now we integrate with respect to x,
so we'll have 16 times x to the third divided by three
minus four times x to the fourth divided by four.
Let's go and simplify this and we have
16/3 x cubed minus x to the fourth.
So when x is two, we have 16/3 times two to the third
minus two to the fourth,
and then minus, notice when x is zero,
these will both be zero,
and this difference comes out to 80/3,
which will be approximately 26.6667.
Again, this is charge, so this would be coulombs.
So again, this would be the total charge over this region
given the density function,
sigma of x comma y equals two x y.
I hope you found this helpful.