﻿ ﻿POLAR AREA (BC TOPIC ONLY) - Applications of Integration - AP CALCULUS AB & BC REVIEW - Master AP Calculus AB & BC

## Master AP Calculus AB & BC

Part II. AP CALCULUS AB & BC REVIEW

CHAPTER 9. Applications of Integration

POLAR AREA (BC TOPIC ONLY)

You may have wondered why we didn’t discuss polar arc length. Perhaps you answered that question for yourself. Remember, any polar equation is easily expressed parametrically using the formulas x = rcos θ and y = rsin θ. Therefore, finding polar arc length equates to finding parametric arc length. However, you will need to be able to calculate the area enclosed by polar curves. Smile—this is the last you will see of parametric and polar equations for the AP test. Polar area isn’t hard, but it is different, and it will take a moment or so to get used to.

No self-respecting calculus topic comes without a formula to despair over and ultimately memorize. The formula for polar area is different from all previous area formulas, because it is not based on rectangles. Instead, polar area uses an infinite number of sectors to find area. A sector is a hunk of circle; for example, a piece of pie is a sector of the entire pie. The area of a sector of a circle is given by but since we are using an infinite number of θ’s to calculate exact area, we replace θ with dθ. Therefore, the formula for the area bounded between radial lines θ = a and θ = b is We’ll start with an easy example and dip our toes into the swimming pool. Once you see how good the water feels, you won’t mind diving right in.

Example 10: Show that the area of a quarter circle of radius 2 equals π using polar area.

Solution: A circle is quite easy to express in polar form; a circle of radius 2 is just the equation r = 2. Since we are trying to find the area of a quarter circle, you have to find the area of the circle in one of the quadrants. For simplicity’s sake, let’s choose the first quadrant. Our area formula sums the areas of the little sectors in the first quadrant with the formula Integrate and apply the Fundamental Theorem to get You see, that’s not so bad. Let’s try a shaded-region problem to up the ante a little. To solve it, you’ll have to find the area of the outer portion and subtract the inner. You can use your calculator with no shame on this example.

Example 11: Find the shaded area in the graph below of r = 2 + 3 cos θ. Solution: First of all, we should find where the graph intersects the pole (origin).

2 + 3 cos θ = 0

Set your calculator to rectangular mode for a moment to solve this equation. The answers are θ = 2.300523983 and θ = 3.982661324. This is the meat of the problem. As θ goes from 0 to 2.300523983, the top of the graph is drawn excluding the inner loop. As θ goes from 2.300523983 to 3.982661324, the inner loop is drawn. Finally, from 3.982661324 to 2π, the bottom of the graph is drawn, again excluding the inner loop. Therefore, we can get the area of the requested region by doubling the left area above and then subtracting the middle area. (The rightmost area is exactly the same as the leftmost area, since the equation is symmetric about the x-axis.) Therefore, the answer will be The area is approximately 25.822.

The final polar area problem you can expect from the AP test concerns finding the area bounded by multiple functions. These problems are no more difficult if you draw a graph first and proceed very carefully.

Example 12: Find the area bounded by the graphs r = 3 sin θ and r = 1 + cos θ.

Solution: First, we should determine the value of 0 at which the graphs intersect. This is the important step: the shaded region above is defined by 3 sin θ from θ = 0 to θ = .6435011088; however, from θ = .6435011088 to θ = π, the shaded region is bounded by 1 + cos θ. Therefore, the area will be the sum of those two smaller regions. The total area (please use a calculator!) is approximately 1.521.

EXERCISE 6

Directions: Solve each of the following problems. Decide which is the best of the choices given and indicate your responses in the book.

YOU MAY USE A GRAPHING CALCULATOR FOR PROBLEMS 4 AND 5.

1. Find the area bounded by r = —2sin θ.

2. Find the area in the second quadrant bounded by r = cos θ + sin θ.

3. Find the area enclosed by the graph of r = cos 2θ.

4. Find the area of the region bounded by r = 2 + cos θ and r = 2.

5. Find the area bounded by r = 3 — 2sin θ and r = 4cos θ.

1. The entire graph is drawn from θ = 0 to θ = π. If you calculate the area from 0 to 2π, your answer will be two times too big. Other than that, the setup is very easy: This is similar to Example 10, since both problems boil down to finding the area of a circle with radius 1.

2. The radial lines that bound the portion in the second quadrant are θ = π/2 (which makes sense) and θ = 3π/4. The second value makes r = θ (since the sine and cosine of 3π/4 are opposites), which is the intersection at the pole marking the end of the region’s presence in the second quadrant. To find the area, use the formula and multiply the squared term out. Mamma will make a return appearance! That was fantastic—a Mamma substitution and a trigonometric substitution. You can integrate this (using u-substitution for sin 2θ) as follows: It’s not a pretty answer, but calculus ain’t a beauty contest.

3. The best way to approach this problem is to calculate the area of one of the petals and multiply it by 4. This graph is symmetric in just about every possible way, so that makes it a little easier. The petal shaded below is bounded by the radial lines θ = π/4 and θ = 3π/4. Its area is In order to integrate this, you’ll have to use a power-reduction formula. NOTE. The power reduction formula for cos2 2θ = There are 4 petals, so the final answer is 4. The region in common is constructed, as shown in the diagram below. Use your calculator to evaluate and sum the definite integrals to get the approximate area of 9.352.

5. First, find the intersection points of the two graphs, and be very careful—there’s more here than meets the eye. For the sake of not writing decimals until our eyes fall out, we’ll use A and B instead of the gigantic decimal intersection values. What causes trouble here is the circle. It will actually pass through points A and B twice, since it draws its graph completely from θ = 0 to θ = π. So, the circle actually hits A when θ = A and when θ = A + π. Use your calculator to convince yourself that this is true! The circle will hit B when θ = B and when θ = B — π. Below are the integrals that make up the area (there are 3) and a graph of the portion of the shaded region they represent. The only tricky integral is the second one. If you try to represent that area with the integral you will be tracing the entire circle an extra time. Make sure that you take the speed with which the graph draws into account. You need to know the exact values of 0 for that specific graph that bound the region. When you add the three integrals, you get 2.496741051 + 3.494545773 + 2.116617352 ≈ 8.108.

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