# Video: Finding the Volume of the Solid Generated by the Revolution of the Region Bounded by a Given Line around the 𝑥-Axis

Consider the region bounded by the curves 𝑦 = 𝑥 + 4, 𝑦 = 0, 𝑥 = 0, and 𝑥 = 3. Determine the volume of the solid of revolution created by rotating this region about the 𝑥-axis.

03:10

### 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.