Video Transcript
Consider the region bounded by
the curves 𝑦 equals 𝑥 plus four, 𝑦 equals zero, 𝑥 equals zero, and 𝑥 equals
three. Determine the volume of the
solid of revolution created by rotating this region about the 𝑥-axis.
Remember, to find the volume of
a solid of revolution for a region rotated about the 𝑥-axis, we use the
following definition. For the solid that lies between
the vertical lines 𝑥 equals 𝑎 and 𝑥 equals 𝑏, whose cross-sectional area in
the plane through 𝑥 and perpendicular to the 𝑥-axis is 𝐴 of 𝑥 for a
continuous function 𝐴, the volume of this solid is given by the definite
integral between 𝑎 and 𝑏 of 𝐴 of 𝑥 with respect to 𝑥.
So, we’ll begin by defining the
various elements in our question. The solid is bounded by the
vertical lines 𝑥 equals zero and 𝑥 equals three, so we’re going to let 𝑎 be
equal to zero and 𝑏 be equal to three. The region is also bounded by
the curves 𝑦 equals 𝑥 plus four and 𝑦 equals zero. And so, the region itself might
look a little something like this. Rotating this region about the
𝑥-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, 𝐴 of 𝑥, is 𝜋 times 𝑥 plus four all squared. Our volume is therefore equal
to the definite integral between zero and three of 𝜋 times 𝑥 plus four all
squared with respect to 𝑥. 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 𝑥 plus four all squared d𝑥. Now, we have two choices
here. We could use integration by
substitution or we could distribute the parentheses to evaluate our definite
integral. Let’s use integration by
substitution.
We let 𝑢 be equal to 𝑥 plus
four. That’s the inner part of our
composite function. Differentiating 𝑢 with respect
to 𝑥, and we simply get one. Now, whilst d𝑢 by d𝑥 is not a
fraction, we treat it a little like one. And this means we can
alternatively write this as d𝑢 equals d𝑥. We then replace 𝑥 plus four
with 𝑢 and d𝑥 with d𝑢. We are going to need to do
something with our limits, though. We use our substitution; this
is 𝑢 equals 𝑥 plus four. Our lower limit is when 𝑥 is
equal to zero. So, 𝑢 is equal to zero plus
four, which is equal to four. Our upper limit is when 𝑥 is
equal to three. So, 𝑢 is equal to three plus
four, which is seven.
And we’re now ready to evaluate
the volume, 𝜋 times the definite integral between four and seven of 𝑢 squared
with respect to 𝑢. 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 𝑢 cubed
over three between four and seven. That’s 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 𝑥-axis to be 93𝜋 cubic units.
