### Video Transcript

Consider the region bounded by the
curves π¦ equals π₯ plus four, π¦ equals zero, π₯ equals zero, and π₯ equals
three. Determine the volume of the solid
of revolution created by rotating this region about the π₯-axis.

Remember, to find the volume of a
solid of revolution for a region rotated about the π₯-axis, we use the following
definition. For the solid that lies between the
vertical lines π₯ equals π and π₯ equals π, whose cross-sectional area in the
plane through π₯ and perpendicular to the π₯-axis is π΄ of π₯ for a continuous
function π΄, the volume of this solid is given by the definite integral between π
and π of π΄ of π₯ with respect to π₯.

So, weβll begin by defining the
various elements in our question. The solid is bounded by the
vertical lines π₯ equals zero and π₯ equals three, so weβre going to let π be equal
to zero and π be equal to three. The region is also bounded by the
curves π¦ equals π₯ plus four and π¦ equals zero. And so, the region itself might
look a little something like this. Rotating this region about the
π₯-axis, and we obtain a three-dimensional shape as shown. The cross-sectional shape of our
solid will be a circle. And the area of each circle will be
given by π times radius squared.

Now, the radius of each circle will
be the value of the function at that point. And so, the function for its area,
π΄ of π₯, is π times π₯ plus four all squared. Our volume is therefore equal to
the definite integral between zero and three of π times π₯ plus four all squared
with respect to π₯. Since π is a constant, we can take
this outside of our integral and rewrite the volume as π times the definite
integral between zero and three of π₯ plus four all squared dπ₯. Now, we have two choices here. We could use integration by
substitution or we could distribute the parentheses to evaluate our definite
integral. Letβs use integration by
substitution.

We let π’ be equal to π₯ plus
four. Thatβs the inner part of our
composite function. Differentiating π’ with respect to
π₯, and we simply get one. Now, whilst dπ’ by dπ₯ is not a
fraction, we treat it a little like one. And this means we can alternatively
write this as dπ’ equals dπ₯. We then replace π₯ plus four with
π’ and dπ₯ with dπ’. We are going to need to do
something with our limits, though. We use our substitution; this is π’
equals π₯ plus four. Our lower limit is when π₯ is equal
to zero. So, π’ is equal to zero plus four,
which is equal to four. Our upper limit is when π₯ is equal
to three. So, π’ is equal to three plus four,
which is seven.

And weβre now ready to evaluate the
volume, π times the definite integral between four and seven of π’ squared with
respect to π’. We know that we can integrate a
polynomial term whose exponent is not equal to negative one by adding one to the
exponent and then dividing by that new value. So, we have π times π’ cubed over
three between four and seven. Thatβs π times seven cubed over
three minus four cubed over three, which becomes π times 279 over three. 279 divided by three is 93. And so, we find the volume obtained
when we rotate our region about the π₯-axis to be 93π cubic units.