We're going to use the washer method
to determine the volume of the solid formed
when the region bounded by y equals two
times the square root of x, y equals zero,
and x equals four is rotated about x equals five,
so because the axis of rotation is the vertical axis,
we'll be using this formula here to determine the volume
where the volume is equal to pi times integral
from c to d of the square root of big r of y
minus the square of little r of y
integrated with respect to y,
so because we have a vertical axis of rotation,
we integrate with respect to y
and the outer radius and inner radius
must both be expressed as a function of y,
so let's first graph the bounded region
and look at the solid when rotating it by x equals five.
So the blue graph is y equals two times square root of x.
This green horizontal line is y equals zero,
and this green vertical line is x equals four.
So this yellow region is the bounded region,
which we'll rotate about x equals five,
this axis here, notice how it would give us
this volume on the right, so our goal
is to find this volume using the washer method.
Before we set up the interval,
notice how the vertical on x equals four
intersects the graph y equals two times the square root of x
at the point four comma four, so for this bounded region,
the interval for x would be from zero to four
and the interval for y would also be
from zero to four, and we'll used a closed interval
from zero to four for y for our limited integration
because we are going to integrate with respect to y.
And it's always helpful to sketch
a representative rectangle to help
us set up the integral.
We'll use this representative rectangle
shown here in red, where if we were
to rotate this rectangle about the axis of rotation,
it would give us one washer volume of the solid.
Whenever using the disc or washer method,
we sketch the rectangle perpendicular
to the axis of rotation, and notice how
the thickness of this rectangle
would be delta y, once again verifying
we integrate with respect to y.
And now let's work on determining big r of y
and little r of y.
Big r of y is the outer radius,
which should be the distance from
the axis of rotation to the corresponding point
on the blue graph, which would be this distance here.
This is big r of y, notice how big r of y
is controlled by y equals two times the square root of x,
so because we need big r of y to be a function of y,
we'll have to solve this equation for x.
Let's do that now.
So if y is equal to two times the square root of x
and we divide both sides by two,
we can say the square root of x
is equal to y divided by two.
Now we can undo the square root by squaring both sides,
which would give us the equation x equals y squared
divided by two squared of y squared divided by four,
so this equation here is equivalent
to the original equation as long as we're assuming
y is greater than or equal to zero.
Now we can use this to determine big r of y.
Notice how the distance from the axis of rotation
to the y axis would be five units,
so big r of y would actually be five units
minus this horizontal distance here,
controlled by the function y equals two
times the square root of x, which is the same
as x equals y squared divided by four,
which means big r of y is equal
to five minus y squared divided by four.
Again, this gives us the distance
from the axis of rotation to the y axis
minus the distance from the y axis to the blue function,
which gives us big r of y.
And now little r of y is much easier.
It's the inner radius, and notice how
the distance from this region to x equals five
is a constant, it's always equal to one unit,
where again this would be little r of y
which is just equal to one.
So we have little r of y is equal to one,
now we have all the information we need to find the volume.
The volume, v, is equal to pi times the integral
from zero to four, where this would be
the integral along the y axis of this bounded region,
and then we have the square of big r of y,
which would be the quantity five minus y squared
divided by four squared minus little r of y squared
which would be one squared, integrated with respect to y.
So this integral will give us the volume of the solid.
Let's evaluate this on the next slide.
Let's first square this binomial,
so we have quantity five minus y squared
divided by four times the quantity five
minus y squared divided by four,
so multiplying, we'd have 25, then we'll have
minus 5/4 y squared minus another five fourths y squared.
That would be negative 10/4 y squared
or minus 5/2 y squared, and then finally
we have plus 1/16 y to the fourth.
So we have pi times the integral from zero to four
of, again we have 25 minus 5/2 y to the second
plus 1/16 y to the fourth minus,
of course one squared is just one,
so we only have two like terms,
25 minus one, of course is 24.
So now let's find the anti-derivative.
The integral of 24 with respect of y
would be 24 y, and then here we're going
to have minus 5/2 times y to the third
divided by three, that would be minus 5/6 y
to the third plus here we'll have
1/16 times y to the fifth divided by five,
16 times five is 80, so we'll have 1/80 y to the fifth.
Now we need to find big f of b minus big f of a,
so when y is equal to four, we have 24 times four
minus 5/6 times four to the third
plus 1/80 times four to the fifth,
and then notice when y is zero,
all the terms would be zero.
So simplifying, we have pi times the quantity,
24 times four is equal to 96, 5/6 times four
to the third is equal to 160/3
and 1/80 times four to the fifth
is equal to 64/5, this simplifies
to pi times 832/15 so the exact volume
is 832 pi divided by 15 cubit units.
If we want a decimal approximation using the calculator,
we'd have 832 pi divided by 15,
which would give us approximately 174.2537 cubic units.
So here we have both the exact and the approximate volume
of the solid formed by rotating this yellow region
about x equals five or this solid shown here.
I hope you found this helpful.