Tag Archives: area between curves

Area & Volume (Type 4)

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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.

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 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: A=\tfrac{1}{2}\int\limits_{{{\theta }_{1}}}^{{{\theta }_{2}}}{{{\left( r\left( \theta  \right) \right)}^{2}}}d\theta

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

Next Posts:

Friday March 17: Table and Riemann sums (Type 5)

Tuesday Match 21: Differential Equations (Type 6)

Friday March 24: Others (Type 7: related rates, implicit differentiation, etc.)

Tuesday March 28: for BC Parametric Equation (Type 8)

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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

Area Between Curves

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Applications of Integration 1 – Area Between Curves

The first thing to keep in mind when teaching the applications of integration is Riemann sums. The thing is that when you set up and solve the majority of application problems you cannot help but develop a formula for the situation. Students think formulas are handy and go about memorizing them badly. By which I mean they forget or never learn where the various things in the formula come from. A slight change in the situation and they are lost. Behind every definite integral stands a Riemann sum; each application should be approached through its Riemann sum. If students understand that, they will make fewer mistakes with the “formula.”

As I suggested in a previous post, I believe all area problems should be treated as the area between two curves. If you build the Riemann sum rectangle between the graph and the axis and calculate its vertical side as the upper function minus the lower (or right minus left if you use horizontal rectangles) you will always get the correct integral for the area. If the upper curve is the x-axis, then the vertical sides of the Riemann sums are (0 – f(x)) and you get a positive area as you should.

If both your curves are above the x-axis, then it is tempting to explain what you are doing as subtracting the area between the lower curve and the x-axis from the area between the upper curve and the x-axis. And this is not wrong. It just does not work very smoothly when one, both or parts of either are below the x-axis. Then you go into all kinds of contortions explaining things in terms of positive and negative areas.  Why go there?

Regardless of where the two curves are relative to the x-axis, the vertical distance between them is the upper value minus the lower, f(x) – g(x). It does not matter if one or both functions are negative on all or part of the interval, the difference is positive and the area between them is

\displaystyle \underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{i=1}^{n}{\left( f\left( {{x}_{i}} \right)-g\left( {{x}_{i}} \right) \right)\Delta {{x}_{i}}}=\int_{a}^{b}{f\left( x \right)-g\left( x \right)dx}.

Furthermore, this Riemann sum rectangle is used in other applications. It is the one rotated in both the washer and shell method of finding volumes. So in area and all applications be sure your students don’t just memorize formulas, but keep their eyes on the rectangle and the Riemann sum.

Finally, if the graphs cross in the interval so that the upper and lower curves change place, you may (1) either break the problem into several pieces so that your integrands are always of the form upper minus lower, or (2) if you intend to do the computation using technology, set up the integral as

\displaystyle \int_{a}^{b}{\left| f\left( x \right)-g\left( x \right) \right|dx}.