We want to find the mass of a lamina
bounded by y equals 2 times the square root of x,
x equals nine and y equals zero with the given
density function here.
So the lamina would be bounded by these three equations here
in the x-y plane.
We notice how this top curve is y equals 2 times
the square root of x.
This horizontal line is y equals zero
and this vertical line is x equals nine.
This will also determine the coordinates
of the corner points here.
We have the origin, which will be 0 comma 0.
Here we have the point nine comma zero.
Here we have the point nine comma six.
This is a fairly straightforward question
because if ro of x comma y is a continuous density function,
of a lamina in the plan region R,
then the mass of the lamina is equal to the double integral
over the region R of the density function.
So in this case we have the mass m is equal to
the double integral.
The end function is going to be sigma of x comma y,
which is x plus one.
Now the differential A will have either dxdy or dydx
and we'll set up both of the double integrals
for this example.
Let's begin with dxdy, meaning we integrate
with respect to x first.
Then with respect to y.
For we integrate with respect to x first,
we need to determine the limits of integration for x
by determining how this area is bounded to the left
and to the right.
We notice how the left is bounded by this curve
given by y equals two times the square root of x.
And to the right, it's bounded by x equals nine.
We need to solve y equals 2 times the square root of x
for x, if you want limits of integration for x.
So starting with y equals two times the square root of x,
we would divide both sides by two,
giving us the square root of x equals y divided by two.
And then we'd square both sides of the equation
given us x equals y squared divided by four.
So y squared divided by four would be the lower limit
of integration with respect to x.
And then to the right, again it's bounded by x equals nine,
so nine is the upper limit of integration with respect to x.
Now with respect to y, because we already have the area
bound to the left and right,
for y we integrate from zero all the way up
to this high point, where y equals six.
So limits of integration with respect to y
would be from zero to six.
Now let's also set this up using
the other order of integration.
So we have the mass would also be equal to
the double integral of x plus one,
but not when you have respect to y first,
then with respect to x.
So because you make it respect to y first,
it would determine which functions bound this area
on the bottom and on top.
Notice how y equals zero bounds the area on the bottom
and y equals 2 times the square root of x
bounds the area on the top.
Which means with respect to y, we integrate from zero
all the way to 2 times the square root of x.
Now the area's bounded on the bottom and on the top.
We integrated with respect to x from zero
all the way out to x equals nine.
So with respect to x, we integrate from zero to nine.
Let's go and use this formula of the integral
in order to determine the mass.
When we integrate with respect to y,
we treat x as a constant and therefore
the antiderivative with respect to y,
which is the x plus one times y...
Again we're treating x and one as a constant.
And now we perform substitution for y.
And because we're treating x plus y as a constant,
we have the integrals from zero to nine of
the constant x plus one times when y equals two
square root of x, we have to two square root of x.
When y is zero, of course we have zero.
Let's go ahead and write this as
the integral from zero to nine
of, let's write square root of x is x to the one half,
so we have 2x to the one-half times
the quantity x plus one, as in a grand function,
let's go and distribute here.
This is x to the first.
So we got the integral from zero to nine.
We'd have 2x to the one half, plus one,
is three halves.
Two x to the three halves, plus 2x the one half.
And now we integrate with respect to x.
So we'd have two times x to the three halves, plus one,
that's x to the five halves.
Instead of dividing by five halves,
that's multiplied by two fifths.
And then plus two times x to the one half, plus one,
is three halves.
Instead of dividing by three halves,
let's multiply by two thirds.
Let's go head and simplify.
So here we have four fifths
times x to the five halves.
Plus four thirds, x to the three halves.
So when x is nine, we have four fifths
times nine to the five halves, plus four thirds
times nine to the three halves,
minus...
Notice, when x is zero both of these would be zero.
So the exact mass would be four fifths
times nine to the five halves
plus four thirds, times nine to the three halves.
Now we could factor this or change the form.
Let's go and leave it in this form though.
I prefer it to a decimal.
So going into the calculator,
notice how the decimal exact value is
230.4.
Now because we're not given units though,
we don't know if this is slugs or kilograms,
but this would be the mass of the lamina
over the region R, with the given density function.
I hope you found this helpful.