### Video Transcript

A nonuniform square lamina,
π΄π΅πΆπ·, has a side length of 15 inches. The density of the lamina across
its area from its edge π΄π΅ is given by the function π of π₯ is equal to one plus
the sin of two plus 0.02π₯ squared pounds per square inch, where π₯ is the distance
from the edge π΄π΅ of the lamina. What is the mass, to the nearest
pound, of the rectangle remaining after removing the part of the lamina which is
greater than six inches from the edge π΄π΅?

This question refers to a
nonuniform square lamina π΄π΅πΆπ·, which has a side length of 15 inches. A lamina is a two-dimensional
planar surface with mass π and surface density π, where π may be a constant, a
function of π₯, or a function of π¦, or a function of π₯ and π¦, in units of mass
per area squared, in our case, inches squared. A lamina is said to be uniform if
the surface density π is constant or nonuniform if π is a function of π₯, π¦, or
π₯ and π¦. For a uniform lamina, the mass π
is π, which is the surface density, times the area. And for a nonuniform lamina, the
mass π is the integral over the area of the surface density function π.

In our case, we have a nonuniform
square lamina, π΄π΅πΆπ·, which has side length 15 inches. The density of the lamina across
its area from its edge π΄π΅ is given by the function π of π₯ is equal to one plus
the sin of two plus 0.02π₯ squared pounds per square inch. π₯ in our function is the distance
from the edge π΄π΅ of the lamina. Placing our lamina in the
π₯π¦-plane, weβd like to find the mass of the rectangle remaining after removing the
part of the lamina, which is greater than six inches from the edge π΄π΅. And remember that our mass is the
integral of the function π over the area. Our function π depends on π₯
only. And we know our length in the
π¦-direction is a constant, 15. This means we can multiply our
integral by 15 and integrate with respect to π₯ only. This gives us the mass is 15 times
the integral of our function π of π₯ with respect to π₯.

Now, we need to consider our
boundaries for π₯ β that is the limit of integration. As weβre dealing with a physical
object, the density must be positive. So, our square will be in the first
quadrant. And the simplest place to put this
is with one corner at the origin. Since our rectangle has a side
length of six, if our lower limit of Integration is zero, our upper limit is
six. Weβll come back and take another
look at this once weβve done our integration and found the mass. So letβs do that now. Our mass is 15 times the integral
between zero and six of π of π₯ with respect to π₯, which is 15 times the integral
between zero and six of one plus sin two plus 0.02π₯ squared with respect to π₯. This is not simple to integrate by
hand. But we can input this definite
integral into a calculator. This gives us a mass of 159.034
pounds, which to the nearest pound is 159.

Letβs just go back and consider our
π₯ boundaries again. Since our function is a function of
π₯ only, our π¦ is a constant. So, the square can be placed
anywhere we like in the vertical direction. To make things simple in the
π₯-direction, we chose π₯ is equal to zero as our lower bound. But we could have put our square
anywhere along the π₯-axis, say at π₯ is equal to π. The value weβre putting into our
function π is the distance from the side π΄π΅ in the π₯-direction. So, if the side π΄π΅ is at π₯ is
equal to π, then at π₯ is equal to π₯ π, for example, the distance from our side
π΄π΅, which is our input, is π₯ π minus π. So, if the side π΄π΅ is at π₯ is
equal to π, we would integrate the function π of π₯ minus π between π and π
plus six.

When we did our calculation, we
chose π equal to zero. And this gave us a mass of 15 times
the integral between zero and six of π of π₯ with respect to π₯. Our mass was, therefore, 159
pounds.