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Consider the region bounded by the curves π¦ equals π₯ plus four, π¦ equals zero, π₯ equals zero, and π₯ equals three.
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Determine the volume of the solid of revolution created by rotating this region about the π₯-axis.
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Remember, to find the volume of a solid of revolution for a region rotated about the π₯-axis, we use the following definition.
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For the solid that lies between the vertical lines π₯ equals π and π₯ equals π, whose cross-sectional area in the plane through π₯ and perpendicular to the π₯-axis is π΄ of π₯ for a continuous function π΄, the volume of this solid is given by the definite integral between π and π of π΄ of π₯ with respect to π₯.
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So, weβll begin by defining the various elements in our question.
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The solid is bounded by the vertical lines π₯ equals zero and π₯ equals three, so weβre going to let π be equal to zero and π be equal to three.
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The region is also bounded by the curves π¦ equals π₯ plus four and π¦ equals zero.
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And so, the region itself might look a little something like this.
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Rotating this region about the π₯-axis, and we obtain a three-dimensional shape as shown.
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The cross-sectional shape of our solid will be a circle.
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And the area of each circle will be given by π times radius squared.
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Now, the radius of each circle will be the value of the function at that point.
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And so, the function for its area, π΄ of π₯, is π times π₯ plus four all squared.
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Our volume is therefore equal to the definite integral between zero and three of π times π₯ plus four all squared with respect to π₯.
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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 π₯ plus four all squared dπ₯.
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Now, we have two choices here.
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We could use integration by substitution or we could distribute the parentheses to evaluate our definite integral.
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Letβs use integration by substitution.
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We let π’ be equal to π₯ plus four.
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Thatβs the inner part of our composite function.
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Differentiating π’ with respect to π₯, and we simply get one.
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Now, whilst dπ’ by dπ₯ is not a fraction, we treat it a little like one.
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And this means we can alternatively write this as dπ’ equals dπ₯.
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We then replace π₯ plus four with π’ and dπ₯ with dπ’.
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We are going to need to do something with our limits, though.
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We use our substitution; this is π’ equals π₯ plus four.
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Our lower limit is when π₯ is equal to zero.
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So, π’ is equal to zero plus four, which is equal to four.
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Our upper limit is when π₯ is equal to three.
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So, π’ is equal to three plus four, which is seven.
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And weβre now ready to evaluate the volume, π times the definite integral between four and seven of π’ squared with respect to π’.
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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.
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So, we have π times π’ cubed over three between four and seven.
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Thatβs π times seven cubed over three minus four cubed over three, which becomes π times 279 over three.
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279 divided by three is 93.
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And so, we find the volume obtained when we rotate our region about the π₯-axis to be 93π cubic units.