WEBVTT
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Use the slicing method to find the volume of the solid whose base is the region bounded by the lines.
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π₯ plus five π¦ is equal to five, π₯ is equal to zero, and π¦ is equal to zero.
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If the cross sections taken perpendicular to the π₯-axis are semicircles.
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The question is asking us to use the slicing method to find the volume of this solid.
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And we recall we can use the slicing method in the following four steps.
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First, we need to determine the region which represents the base of our solid.
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And We can do this by sketching the information given to us in the question.
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Next, we need to determine the shape and orientation of the cross sections of our solid.
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And again, we can do this by sketching the information given to us in the question.
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Third, we need to find the area of our cross sections.
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And if our cross sections are taken perpendicular to the π₯-axis, we want our area in terms of π₯.
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If our cross sections are perpendicular to the π¦-axis, we want our area in terms of π¦.
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And if our cross sections are perpendicular to the π§-axis, we want our area in terms of π§.
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Finally, we just integrate our formula for the area over an appropriate interval.
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And then this gives us the volume of our solid.
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Letβs start by finding the base of our solid.
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The question tells us the base of our solid is bound by the lines π₯ plus five π¦ is equal to five, π₯ is equal to zero, and π¦ is equal to zero.
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Since weβre told these are lines, this tells us that our base lies within the π₯π¦-plane.
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So letβs sketch a graph of our base.
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We know the lines π₯ is equal to zero and π¦ is equal to zero coincide with our axes.
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And we can rearrange the straight line π₯ plus five π¦ is equal to five to give us π¦ is equal to negative π₯ over five plus one.
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And since this is a standard form for a straight line, we know that the slope of this straight line is negative one-fifth.
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And the π¦-intercept is one.
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So we can sketch a graph of π₯ plus five π¦ is equal to five.
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We know it intercepts the π¦-axis when π¦ is equal to one.
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And we can also find the π₯-intercept.
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This will be when π₯ is equal to five.
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Since the π₯-intercept will be when π¦ is equal to zero.
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Now since our base is bounded by the lines π₯ is equal to zero, π¦ is equal to zero, and π₯ plus five π¦ is equal to five, we can see that we found the sketch of our base.
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In fact, itβs a right-angled triangle with height one and width five.
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So we found that our base is a right-angled triangle.
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Next, we need to determine the shape of our cross sections.
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And the question tells us that any cross section taken perpendicular to the π₯-axis will be a semicircle.
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So letβs use this information to sketch our cross sections.
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Weβll start by sketching our base on a set of three-dimensional axes.
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Next, weβre told that the cross sections taken perpendicular to the π₯-axis are semicircles.
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So weβll draw a line which is perpendicular to the π₯-axis.
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And this cross section will be a semicircle.
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In fact, we can also see the diameter of our semicircle will be the height of our perpendicular line.
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So we now have that our cross sections are semicircles perpendicular to the π₯-axis with a diameter equal to the height of our perpendicular line.
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We now need to calculate the area of our cross section.
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And remember, since weβre taking our cross sections perpendicular to the π₯-axis, we want our formula in terms of π₯.
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And we recall the area of a circle with radius π is ππ squared.
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So the area of a semicircle will be ππ squared all divided by two.
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Now to calculate the area of our cross sections, weβll go back to the sketch of our base.
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And we know the diameter of a cross section taken at π₯ will be the length of this vertical line.
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And the length of this vertical line will just be its π¦-coordinate.
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And we found that its π¦-coordinate is equal to negative π₯ divided by five plus one.
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So our cross sections have a diameter of negative π₯ divided by five plus one.
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And if we divide this by two, weβll find the radius of our cross section.
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So we can calculate the area of our cross section by using a half ππ squared.
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This gives us π over two multiplied by a half multiplied by negative π₯ over five plus one squared.
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Then distributing our exponent over our parentheses and simplifying, we have π over eight multiplied by π₯ squared over 25 minus two π₯ over five plus one.
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So weβve now found a formula for the area of our cross section.
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And since our cross section is given perpendicular to the π₯-axis, weβve written our area in terms of π₯.
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Finally, we need to integrate our area over an appropriate interval to find the volume of our solid.
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We can see from our sketch that the values of π₯ vary from zero to five.
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This tells us we can calculate the volume of our solid as the integral from zero to five of the area of our cross sections perpendicular to the π₯-axis with respect to π₯.
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So by using the formula we found for the areas of our cross section, we have that our volume is equal to the integral from zero to five of π over eight times π₯ squared over 25 minus two π₯ over five plus one with respect to π₯.
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First, weβre going to take the constant coefficient of π over eight outside of our integral.
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Then weβre going to integrate each term by using the power rule for integration.
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This gives us π over eight multiplied by π₯ cubed over 75 minus π₯ squared over five plus π₯ evaluated at the limits of our integral.
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π₯ is equal to zero and π₯ is equal to five.
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Evaluating at the upper limit of our integral gives us five cubed over 75 minus five squared over five plus five.
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We then need to subtract the evaluation at the lower limit of our integral.
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However, we see that all three terms share a factor of π₯.
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So when we evaluate this at π₯ is equal to zero, we will get zero.
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Evaluating this expression gives us π over eight multiplied by five over three, which simplifies to give us five π over 24.
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Therefore, by using the slicing method, we found the volume of the solid whose base is the region bounded by the lines.
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π₯ plus five π¦ equals five, π₯ is equal to zero, and π¦ is equal to zero.
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With cross sections perpendicular to the π₯-axis are semicircles is equal to five π divided by 24.