Welcome to an example
on how to determine the volume of a solid by slices
or with a known cross section.
A volume is described as follows:
The base is the region bounded by
y equals six minus 2/3 x squared
and y equals zero.
And, therefore, the base would be this bounded region here
in the xy-plane,
where the region is bounded by the parabola
and the x-axis.
Every cross section parallel to the x-axis
is a triangle whose height and base are equal.
And so we're trying to find the volume of this solid here,
where, again, this region here is the base
shown at the bottom
and each cross section parallel to the x-axis
is a triangle where the base is equal to the height.
In this case, because the cross sections
are parallel to the x-axis,
we actually will not be using this volume formula here,
we'll be using the formula the volume V
is equal to the integral from c to d
of A of y
integrated with respect to y.
Here, we integrate with respect to y
because each cross section is parallel to the x-axis,
and, therefore, we'll be integrating from zero to six
along the y-axis
and A of y is going to be a function of y
that gives the area of any cross-sectional face.
Looking at our bounded region here again, if we needed to,
we could find the x-intercepts by setting
y equal to zero and solving for x,
and to find the y-intercept,
we could set x equal to zero and solve for y.
To begin to determine A of y
which, again, gives us the area of any cross-sectional face,
let's consider one segment in the base
parallel to the x-axis.
Let's say this segment here.
Notice how this segment would correspond to the length
of the base of any cross-sectional triangle.
So let's begin by expressing this base as a function of y.
Notice how it's controlled by
y equals six minus 2/3 x squared.
So for our first step, we need to solve
this equation for x.
So we have y equals six minus 2/3 x squared.
Let's begin by adding 2/3 x squared to both sides
as well as subtracting y on both sides.
So we'd have 2/3 x squared
equals six minus y.
Now, let's multiply both sides by the reciprocal of 2/3,
which would be 3/2.
So we'd have x squared equals 3/2
times the quantity six minus y.
Distributing, we have x squared equals nine minus 3/2 y.
Now we'll take the square root
of both sides of the equation.
Of course, we are going to have a plus or minus here.
So we have x equals plus or minus the square root
of the quantity nine minus 3/2 y.
So looking in the segment again, notice how
the point on the right would have an x-coordinate
that's positive square root
of the quantity nine minus 3/2 y.
And on the left, the x-coordinate would have to be negative
square root of the quantity nine minus 3/2 y.
So knowing this information,
we should now be able to express the entire length
as a function of y.
Since half the length would be positive
square root of the quantity nine minus 3/2 y,
that means the entire length, this length here,
would have to be twice as longer,
two times the square root of
the quantity nine minus 3/2 y.
So this would be the length of the base of the triangle
as a function of y.
Let's go ahead and label that.
This length would be two times the square root
of the quantity nine minus 3/2 y.
And because the base and height have the same length,
we now also know the height
would have a length of two times the square root
of the quantity nine minus 3/2 y.
And this is all the information we need in order to find
our integrand function A of y.
A of y
is going to be equal to 1/2 times the base times the height,
which in this case would be 1/2 times,
the base times the height,
so we'd have two times the square root of the quantity
nine minus 3/2 y
times another two
times the square root of the quantity nine minus 3/2 y.
So multiplying,
notice how we have A of y equal,
well, 1/2 times two times two is two.
When we multiply the two square roots, we just get
the quantity nine minus 3/2 y.
So A of y is equal to 18 minus three y.
So this is the hard part of our question.
Now we can set up the integral to determine the volume.
The volume of the solid is equal to the integral,
again, we're integrating with respect to y,
so integrate along the y-axis from zero
to the y-intercept of the parabola which is six.
So we integrate from zero to six
of A of y which is 18 minus three y.
Let's go ahead and evaluate this on the next slide.
Integrating with respect to y, we'd have
18 y minus three times
y squared divided by two or minus 3/2 y squared.
Now, we need to find big F of b minus big F of a.
So when y is equal to six, we'd have 18 times six
minus 3/2 times six squared
minus,
when y is zero, notice both terms would be zero.
So simplifying, we have 18 times six, that's 108,
minus,
here we have 3/2 times 36, that's 54,
which gives us 54.
So this is the volume of the solid,
and, therefore, this would be cubic units.
So, again, this is the volume of the solid
where the base would be this bounded region here
and each cross section parallel to x-axis
is a triangle where the base and the height are equal.
I hope you found this helpful.