Video: EG19M1-DiffAndInt-Q08

If a region bounded by the two curves 𝑦 = 6 βˆ’ π‘₯ and 𝑦 = √π‘₯ and the π‘₯-axis is revolved completely about the π‘₯-axis, find the volume of the solid generated.

07:49

Video Transcript

If a region bounded by the two curves 𝑦 equals six minus π‘₯ and 𝑦 equals the square root of π‘₯ and the π‘₯-axis is revolved completely about the π‘₯-axis, find the volume of the solid generated.

Before we go forward, let’s sketch out this image. Here’s the π‘₯- and 𝑦-axis. 𝑦 equals the square root of π‘₯ would look something like this. 𝑦 equals six minus π‘₯ would look something like this. And the figure is bounded by the π‘₯-axis. The solid generated by revolving this region around the π‘₯-axis would look something like this. To find the volume of something like this, we take the definite integral of πœ‹π‘Ÿ squared dπ‘₯, where π‘Ÿ is the vertical distance from the axis of rotation. And that means that π‘Ÿ is a function of π‘₯. But our region is bounded by two different functions. We need to find the volume of the blue piece by taking a definite integral of πœ‹ times the square root of π‘₯ squared. And we’ll add that to the volume created by six minus π‘₯. But how do we decide and define these definite integrals?

We know that our smallest point of the square root of π‘₯ function is zero. To find out the other two parts of the definite integral, we’ll need to find the intersection of 𝑦 equals six minus π‘₯ and 𝑦 equals the square root of π‘₯. We can find that intersection by setting these two functions equal to each other: the square root of π‘₯ equals six minus π‘₯. To get rid of that square root, we square both sides and then we have π‘₯ equals six minus π‘₯ squared. To solve, we need to expand. Six minus π‘₯ times six minus π‘₯ 36 minus six π‘₯ minus six π‘₯ plus π‘₯ squared is equal to π‘₯. We can combine like terms. Negative six π‘₯ plus negative six π‘₯ equals negative 12π‘₯. We just rearrange the function to say π‘₯ squared minus 12π‘₯ plus 36. Bring down that π‘₯.

To set this equation equal to zero, we subtract π‘₯ from both sides. We see that zero equals π‘₯ squared minus 13π‘₯ plus 36 we can solve for π‘₯ by factoring. Two factors of 36 that add together to equal negative 13 are negative nine and negative four. So we’ll set π‘₯ minus nine and π‘₯ minus four equal to zero, like this. And we have π‘₯ equals nine and π‘₯ equals four. But we only have one intersection here. So we need to see which of these π‘₯-values is real. We’ll have either 𝑦 equals six minus four or 𝑦 equals six minus nine. When π‘₯ equals four, 𝑦 equals two. And when π‘₯ equals nine, 𝑦 equals negative three. The solution four, two is the solution we’re looking for. It’s the one in the first quadrant. And we can ignore that second π‘₯ equals nine. But what is this telling us?

This dotted line at π‘₯ equals four helps us to define our definite integrals. We’re using the blue function, the square root of π‘₯ from zero to four. And at four, we switch over to the yellow function. The final thing we need to consider is the place where the function 𝑦 equals six minus π‘₯ intersects the π‘₯-axis. 𝑦 equals six minus π‘₯ intersects the π‘₯-axis when 𝑦 equals zero. So we set 𝑦 equal to zero and solve for π‘₯. Subtract six from both sides. Negative six equals negative π‘₯ and π‘₯ equals six. And that means the yellow function is bounded between four and six.

Now, we need to clear some space to find these integrals. We’ll start with the blue piece. We know that πœ‹ is a constant. So we can go ahead and take it out. And the square root of π‘₯ squared equals π‘₯. We need to take πœ‹ times the definite integral from zero to four of π‘₯ dπ‘₯. We’ll need the antiderivative of π‘₯ dπ‘₯ from four to zero. And we’re gonna multiply that by πœ‹.

Moving on to the yellow piece, the six minus 𝑦 function, we can go ahead and pull out our constant. And then, we need to consider what six minus π‘₯ squared would be. Fortunately, we already squared six minus π‘₯ we know that it is π‘₯ squared minus 12π‘₯ plus 36 dπ‘₯. And now, we need the antiderivative of π‘₯ squared minus 12 π‘₯ plus 36. One-third π‘₯ cubed minus 12 divided by two π‘₯ squared. 12 divided by two equals six plus 36 π‘₯ from four to six. And then, we’ll multiply all of that by πœ‹. Notice that both of these terms have a factor of πœ‹. So we can pull that out and save it for the end.

Now, we’re gonna substitute what we know. One-half times four squared minus one-half times zero squared. We know that zero squared term drops out plus one-third times six cubed minus six times six squared plus 36 times six minus one-third times four cubed minus six times four squared plus 36 times four. Four squared equals 16 times one-half equals eight. Six cubed equals 216 divided by three equals 72. Six times six squared is the same thing as six cubed, 216. 36 times six is the same thing as six cubed, 216, minus four cubed is 64 divided by three is 64 thirds. Four squared equals 16 times six equals 96 and 36 times four equals 144. And remember all of this is being multiplied by πœ‹. 72 minus 216 plus 216 equals 72. Negative 96 plus 144 equals 48. 64 thirds plus 48 equals 69 and one-third. 72 minus 69 and one-third equals two and two-thirds, which we can rewrite as eight-thirds. We want to add eight and eight-thirds together. We can rewrite eight as 24 thirds. That way we can add the numerators together. 24 plus eight equals 32. We have 32 thirds that we need to multiply by πœ‹, 32πœ‹ over three units cubed since we’re dealing with volume.

This region bounded by the curves 𝑦 equals six minus π‘₯ and 𝑦 equals the square root of π‘₯ has a volume of 32πœ‹ over three units cubed.

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