Tag Archives: disk

Type 4 Questions: Area and Volume Problems

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Given equations that define a region in the plane students are asked to find its area, the volume of the solid formed when the region is revolved around a line, and/or the region is used as a base of a solid with regular cross-sections. This standard application of the integral has appeared every year since year one (1969) on the AB exam and almost every year on the BC exam. You can be pretty sure that if a free-response question on areas and volumes does not appear, the topic will be tested on the multiple-choice section.

What students should be able to do:

  • Find the intersection(s) of the graphs and use them as limits of integration (calculator equation solving). Write the equation followed by the solution; showing work is not required. Usually no credit is earned until the solution is used in context (as a limit of integration). Students should know how to store and recall these values to save time and avoid copy errors.
  • Find the area of the region between the graph and the x-axis or between two graphs.
  • Find the volume when the region is revolved around a line, not necessarily an axis or an edge of the region, by the disk/washer method.
  • The cylindrical shell method will never be necessary for a question on the AP exams, but is eligible for full credit if properly used.
  • Find the volume of a solid with regular cross-sections whose base is the region between the curves. For an interesting variation on this idea see 2009 AB 4(b)
  • Find the equation of a vertical line that divides the region in half (area or volume). This involves setting up an integral equation where the limit is the variable for which the equation is solved.
  • For BC only – find the area of a region bounded by polar curves: A=\tfrac{1}{2}\int\limits_{{{\theta }_{1}}}^{{{\theta }_{2}}}{{{\left( r\left( \theta  \right) \right)}^{2}}}d\theta
  • For BC only – Find perimeter using arc length integral

If this question appears on the calculator active section, it is expected that the definite integrals will be evaluated on a calculator. Students should write the definite integral with limits on their paper and put its value after it. It is not required to give the antiderivative and if a student gives an incorrect antiderivative they will lose credit even if the final answer is (somehow) correct.

There is a calculator program available that will give the set-up and not just the answer so recently this question has been on the no calculator allowed section. (The good news is that in this case the integrals will be easy or they will be set-up-but-do-not-integrate questions.)

Occasionally, other type questions have been included as a part of this question. See 2016 AB5/BC5 which included an average value question and a related rate question along with finding the volume.

Shorter questions on this concept appear in the multiple-choice sections. As always, look over as many questions of this kind from past exams as you can find.

For some previous posts on this subject see January 9, 11, 2013 and “Subtract the Hole from the Whole” of December 6, 2016

Free-response questions:

  • 2014 AB 2, 2013 AB 5.
  • 2015 AB 2
  • Variations: 2009 AB 4,
  • 2016 AB5/BC5,
  • 2017 AB 1 (using a table),
  • Perimeter 2011 BC 3 and 2014 BC 5

Multiple-choice questions from non-secure exams:

  • 2008 AB 83 (Use absolute value),
  • 2012 AB 10, 92
  • 2012 BC 87, 92 (Polar area)

 

 

 

 

 

 

Revised to add perimeter question 3-16-18,

Revised March 12, 2021

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Challenge

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A student was asked to find the volume of the bowl-shaped figure generated when the curve y = x2 from x = 0 to x = 2 is revolved around the y-axis. She used the disk method and found the volume to be \displaystyle \int_{0}^{4}{\pi ydy}=8\pi . To check her work she used the method of cylindrical shells and found the same answer: \displaystyle\int_{0}^{2}{2\pi x\left( 4-{{x}^{2}} \right)dx}=8\pi .

The second part of the question asked for the value of x for which the bowl would be half full. So she first solved the equation \displaystyle \int_{0}^{k}{\pi ydy=\frac{1}{2}\cdot 8\pi }  and found k=2\sqrt{2}. This is a y-value so the corresponding x-value is \sqrt{2\sqrt{2}}\approx 1.682. She again checked her work by shells by solving \displaystyle \int_{0}^{k}{2\pi x\left( 4-{{x}^{2}} \right)dx=\frac{1}{2}\cdot 8\pi } and found that k=\sqrt{-2\left( \sqrt{2}-2 \right)}\approx 1.082. (This is the x-value.)

Both computations are correct. Can you explain to her why her answers are different? Use the comment box below to share your explanations. I will post mine in a week or so.

Area and Volume Questions

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AP Type Questions 4

Given equations that define a region in the plane students are asked to find its area and the volume of the solid formed when the region is revolved around a line or used as a base of a solid with regular cross-sections. This standard application of the integral has appeared every year since 1969 on the AB exam and all but one year on the BC exam.

If this appears on the calculator active section, it is expected that the definite integrals will be evaluated on a calculator. Students should write the definite integral with limits on their paper and put its value after it.  It is not required to give the antiderivative and if students give an incorrect antiderivative they will lose credit even if the final answer is (somehow) correct.

There is a calculator program available that will give the set-up and not just the answer so recently this question has been on the no calculator allowed section. (The good news is that the integrals will be easy or they will be set-up but do not integrate questions.)

What students should be able to do:

  • Find the intersection(s) of the graphs and use them as limits of integration (calculator equation solving). Write the equation followed by the solution; showing work is not required. Usually no credit is earned until the solution is used in context (as a limit of integration).
  • Find the area of the region between the graph and the x-axis or between two graphs.
  • Find the volume when the region is revolved around a line, not necessarily an axis, by the disk/washer method. (Shell method is never necessary, but is eligible for full credit if properly used).
  • Find the volume of a solid with regular cross-sections whose base is the region between the curves. But see 2009 AB 4(b)
  • Find the equation of a vertical line that divides the region in half (area or volume). This involves setting up and solving an integral equation where the limit is the variable for which the equation is solved.
  • For BC only – find the area of a region bounded by polar curves:

\displaystyle A=\tfrac{1}{2}{{\int_{{{t}_{1}}}^{{{t}_{2}}}{\left( r\left( t \right) \right)}}^{2}}dt

Shorter questions on this concept appear in the multiple-choice sections. As always, look over as many questions of this kind from past exams as you can find.

For some previous posts on this subject see January 9, 11, 2013