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If a region bounded by the two curves π¦ equals six minus π₯ and π¦ equals the square root of π₯ and the π₯-axis is revolved completely about the π₯-axis, find the volume of the solid generated.
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Before we go forward, letβs sketch out this image.
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Hereβs the π₯- and π¦-axis.
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π¦ equals the square root of π₯ would look something like this.
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π¦ equals six minus π₯ would look something like this.
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And the figure is bounded by the π₯-axis.
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The solid generated by revolving this region around the π₯-axis would look something like this.
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To find the volume of something like this, we take the definite integral of ππ squared dπ₯, where π is the vertical distance from the axis of rotation.
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And that means that π is a function of π₯.
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But our region is bounded by two different functions.
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We need to find the volume of the blue piece by taking a definite integral of π times the square root of π₯ squared.
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And weβll add that to the volume created by six minus π₯.
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But how do we decide and define these definite integrals?
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We know that our smallest point of the square root of π₯ function is zero.
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To find out the other two parts of the definite integral, weβll need to find the intersection of π¦ equals six minus π₯ and π¦ equals the square root of π₯.
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We can find that intersection by setting these two functions equal to each other: the square root of π₯ equals six minus π₯.
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To get rid of that square root, we square both sides and then we have π₯ equals six minus π₯ squared.
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To solve, we need to expand.
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Six minus π₯ times six minus π₯ 36 minus six π₯ minus six π₯ plus π₯ squared is equal to π₯.
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We can combine like terms.
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Negative six π₯ plus negative six π₯ equals negative 12π₯.
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We just rearrange the function to say π₯ squared minus 12π₯ plus 36.
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Bring down that π₯.
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To set this equation equal to zero, we subtract π₯ from both sides.
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We see that zero equals π₯ squared minus 13π₯ plus 36 we can solve for π₯ by factoring.
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Two factors of 36 that add together to equal negative 13 are negative nine and negative four.
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So weβll set π₯ minus nine and π₯ minus four equal to zero, like this.
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And we have π₯ equals nine and π₯ equals four.
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But we only have one intersection here.
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So we need to see which of these π₯-values is real.
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Weβll have either π¦ equals six minus four or π¦ equals six minus nine.
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When π₯ equals four, π¦ equals two.
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And when π₯ equals nine, π¦ equals negative three.
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The solution four, two is the solution weβre looking for.
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Itβs the one in the first quadrant.
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And we can ignore that second π₯ equals nine.
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But what is this telling us?
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This dotted line at π₯ equals four helps us to define our definite integrals.
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Weβre using the blue function, the square root of π₯ from zero to four.
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And at four, we switch over to the yellow function.
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The final thing we need to consider is the place where the function π¦ equals six minus π₯ intersects the π₯-axis.
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π¦ equals six minus π₯ intersects the π₯-axis when π¦ equals zero.
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So we set π¦ equal to zero and solve for π₯.
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Subtract six from both sides.
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Negative six equals negative π₯ and π₯ equals six.
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And that means the yellow function is bounded between four and six.
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Now, we need to clear some space to find these integrals.
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Weβll start with the blue piece.
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We know that π is a constant.
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So we can go ahead and take it out.
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And the square root of π₯ squared equals π₯.
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We need to take π times the definite integral from zero to four of π₯ dπ₯.
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Weβll need the antiderivative of π₯ dπ₯ from four to zero.
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And weβre gonna multiply that by π.
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Moving on to the yellow piece, the six minus π¦ function, we can go ahead and pull out our constant.
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And then, we need to consider what six minus π₯ squared would be.
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Fortunately, we already squared six minus π₯ we know that it is π₯ squared minus 12π₯ plus 36 dπ₯.
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And now, we need the antiderivative of π₯ squared minus 12 π₯ plus 36.
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One-third π₯ cubed minus 12 divided by two π₯ squared.
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12 divided by two equals six plus 36 π₯ from four to six.
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And then, weβll multiply all of that by π.
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Notice that both of these terms have a factor of π.
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So we can pull that out and save it for the end.
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Now, weβre gonna substitute what we know.
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One-half times four squared minus one-half times zero squared.
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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.
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Four squared equals 16 times one-half equals eight.
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Six cubed equals 216 divided by three equals 72.
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Six times six squared is the same thing as six cubed, 216.
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36 times six is the same thing as six cubed, 216, minus four cubed is 64 divided by three is 64 thirds.
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Four squared equals 16 times six equals 96 and 36 times four equals 144.
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And remember all of this is being multiplied by π.
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72 minus 216 plus 216 equals 72.
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Negative 96 plus 144 equals 48.
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64 thirds plus 48 equals 69 and one-third.
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72 minus 69 and one-third equals two and two-thirds, which we can rewrite as eight-thirds.
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We want to add eight and eight-thirds together.
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We can rewrite eight as 24 thirds.
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That way we can add the numerators together.
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24 plus eight equals 32.
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We have 32 thirds that we need to multiply by π, 32π over three units cubed since weβre dealing with volume.
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This region bounded by the curves π¦ equals six minus π₯ and π¦ equals the square root of π₯ has a volume of 32π over three units cubed.