Video: Finding the Volume of the Solid Generated by the Revolution of the Region Bounded by a Given Line around the π‘₯-Axis

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

03:10

Video Transcript

Consider the region bounded by the curves 𝑦 equals π‘₯ plus four, 𝑦 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‘Ž and π‘₯ equals 𝑏, whose cross-sectional area in the plane through π‘₯ and perpendicular to the π‘₯-axis is 𝐴 of π‘₯ for a continuous function 𝐴, the volume of this solid is given by the definite integral between π‘Ž and 𝑏 of 𝐴 of π‘₯ with respect to π‘₯.

So, we’ll 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 we’re going to let π‘Ž be equal to zero and 𝑏 be equal to three. The region is also bounded by the curves 𝑦 equals π‘₯ plus four and 𝑦 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œ‹ 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 π‘₯, is πœ‹ times π‘₯ plus four all squared. Our volume is therefore equal to the definite integral between zero and three of πœ‹ times π‘₯ plus four all squared with respect to π‘₯. Since πœ‹ is a constant, we can take this outside of our integral and rewrite the volume as πœ‹ 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. Let’s use integration by substitution.

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

And we’re now ready to evaluate the volume, πœ‹ times the definite integral between four and seven of 𝑒 squared with respect to 𝑒. 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œ‹ times 𝑒 cubed over three between four and seven. That’s πœ‹ times seven cubed over three minus four cubed over three, which becomes πœ‹ 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œ‹ cubic units.

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