Let f(x) = {x if 0≤x≤2
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Question

Let f(x) = {x if 0≤x≤2

2x−2 if 2

−2x+18 if 5

Find the volume of the solid formed when the region bounded by the graph of f, the x-axis, and the line x=6 is revolved about the x-axis.

fig

Accepted Answer

Hello. In this video, we are going to consider the following piece-wise function. We want to calculate the volume of the solid generated by revolving the region bounded by the graph of G of X, the X-axis, and the vertical line X is equals 6, about the X-axis. So let's go ahead and start by drawing our boundaries. Now we know that this region is going to be bounded along the X axis. And it is also bounded by the vertical line of X's equal to 6. Now, in this case here, because we're trying to find the volume of the region that is bounded by rotating about the X axis, What we want to go ahead and do is you want to go ahead and create a volume integral that will help us find the solution. Now, the volume in this case is normally defined as pi multiplied by a definite integral from A to B, of a radius squared DX by the DC rule. However, we need to be careful with the setup in this piece-wise problem. Our upper boundary is going to change depending on what section of the graph we are working on. So, for example, the uppermost curve on the region from 0 to 1 is defined as 3 X. The uppermost curve on the boundary from 1 to 4 is defined as 2/3 X plus 7/3, and the boundary, the upper boundary on the interval from 4 to 6 is defined as -5 X plus 7. So what we're going to end up doing is we're going to define the volume as a sum of definite integrals. So, our volume integral is going to equal to pi multiplied by the sum of 3 definite integrals. Now, we're going to create a definite integral for the boundary from 0 to 1. So that is going to be an integral from 0 to 1, and a radius is going to be the topmost curve minus the bottom, which is going to be 3X minus 0 squared. That is just going to give us 3 X squared. Then we will need to add the next region from 1 to 4, and that radius will be defined as 2/3 X plus 7/3 quantity squared. And for the final region, we will have an integral from 4 to 6. Of 1/2 x plus 7 quantity squared. And all of this will be integrated with respect to X. Now, what we're going to do is we're going to individually evaluate each of the integrals in order to find the volume. So let's go ahead and start with the first integral. Now, for the first integral, we have the integral from 0 to 1 of 3 X squared. By simplifying the integran, we can rewrite this integral to be the integral from 0 to 1 of 9 X 2 DX. Then, by taking the antiderivative, this will simplify to be 3 X cubed evaluated from 0 to 1. Then by the fundamental theorem of calculus part two, we will plug in our upper boundary, which is just going to leave us with 3. Minus the value of the lower boundary plugged in, which is 0, and this is going to simplify to 3. So this is the value of the first integral for our volume. Now let's go ahead and evaluate the 2nd integral. Now, for the 2nd integral, the first thing we're going to do is we are going to expand the integrand. This will give us the integral from 1 to 4. Of 4/9. Exque Plus 28 X. 28 divided by 9 X. Plus 49 divided by 9 DX. Then we will go ahead and take the antiderivative of each of these terms. That will leave us with 4 divided by 27 X cubed plus 14 divided by 9 X squared plus 49 divided by 9 X, and this will be evaluated from 1 to 4. Now we will apply the fundamental theorem part 2. This is going to give us the quantity, 4 divided by 27, multiplied by 4 cubed, plus 14 divided by 9. Multiply by 4 squared, plus 49 divided by 9, multiplied by 4. Then we will have to subtract the value of the lower boundary, but if we plug in one into every term, we will just be left with 4 divided by 27, plus 14 divided by 9, plus 49 divided by 9. Then all we need to do is simplify. This will simplify to be 256 plus 672 plus 500. In '88 Divided by 9 Sorry, divided by 27. -4. Plus 42 + 147 divided by 27, and this is going to fully simplify to give us the value of 49. So this is going to be the value of the second integral. Now, for the final integral, what we are going to do is repeat this process of expanding the integrand. So for our 3rd integral, if we expand the integrand, we can rewrite the integral to be the integral from 4 to 6. Of 14th X2 minus 7 X plus 49 DX. Then we will take the antiderivative of each term, and that will leave us with 1/12 X X cubed. Minus 7 halves x squared. Plus 49 X, and this will be evaluated from 4 to 6. Then by using fundamental theorem part two, we will be left with 1/12 multiplied by 6 cubed. Minus 7 halves, multiplied by 6 squared, plus 49 multiplied by 6 minus the value of the lower boundary, which is going to be 1/12 multiplied by force cubed. Minus 7 halves multiplied by 4 squared, plus 49 multiplied by 4. And once we simplify and combine like terms, we will be left with 18 minus 126 plus 294 minus the quantity 16/3 minus 56 plus 196. And simplifying this is going to give us a final value of 122 divided by 3. This is going to be the value of the 3rd integral in our volume. Now what we're going to do is we're just going to put these values back into our volume equation. So volume is going to equal to pi multiplied by the quantity 3 + 49 + 122 divided by 3, and combining like terms and multiplying together is going to give us a total volume of 278 pi divided by 3. And this is going to be the volume of the function of the piece wise function. And the overall solution to this problem is going to be a, So, I hope this video helps you in understanding how to approach this problem, and we will go ahead and see you all in the next video.

Let f(x) = {x   if 0≤x≤2
      2x−2  if 2<x≤5
      −2x+18 if 5<x≤6.

Find the volume of the solid formed when the region bounded by the graph of f, the x-axis, and the line x=6 is revolved about the x-axis.

Key Concepts

Piecewise Functions

Volume of Solids of Revolution

Definite Integration

Watch next

Let f(x) = {x if 0≤x≤2 2x−2 if 2<x≤5 −2x+18 if 5<x≤6. Find the volume of the solid formed when the region bounded by the graph of f, the x-axis, and the line x=6 is revolved about the x-axis.

Key Concepts

Piecewise Functions

Volume of Solids of Revolution

Definite Integration

Watch next

Let f(x) = {x if 0≤x≤2
2x−2 if 2<x≤5
−2x+18 if 5<x≤6.

Find the volume of the solid formed when the region bounded by the graph of f, the x-axis, and the line x=6 is revolved about the x-axis.