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Introduction to Area, Volume, And Arc Length - At A Glance:
Slicing Cold Pizza
First, let's forget about calculus and use our knowledge of fractions to answer the following question.
Sample Problem
What is the area of the shaded region?
Answer.
Think of the shaded region as a piece of cold pizza. The central angle of the slice is
The central angle of the entire pizza is 2π.
This means the piece is this fraction of the whole pizza:
Since the area of the whole pizza is
π(1)^{2} = π,
the area of the piece is
.
We can use this idea to find the area of a piece of cold anchovy pizza for any pizza described by a polar function
r = f(θ).
First we slice the piece into tiny slices. The central angle of a tiny slice is Δ θ.
The angle θ measures the position of a slice by its angle from the positive x-axis, going counterclockwise. If we look just at the slice at position θ, we can pretend this slice is part of a perfectly round pizza with radius f(θ).
The tiny slice would be the fraction
of the whole, perfectly round pizza. So the area of the tiny slice is approximately
Writing this a little more neatly, the π terms cancel out and we're left with
as the area of the tiny slice.
If we want to add up the areas of all the tiny slices and take the limit as the number of slices approaches ∞ to get an integral, we need to know the reasonable values of θ.
The original piece of cold pizza must go from some angle α to some angle β.
Using those values α and β as the limits of integration, the area of the original piece of pizza is
This is usually written
Our cold, leftover slice of pizza came in handy for more than just satisfying our hunger pains at 4 am. We used the cold slice to derive the area for a slice-like portion of a circle in polar coordinates.
When doing problems with polar coordinates, we can use the formula
All we have to do is put in the correct values for α and β and the correct function for f(θ).
We promised before, when talking about triangles, that you would have the power to derive any area equation you needed. If you forget this formula, don't panic. Just recreate it. Remember that each tiny slice is like a fraction of circle, so the area of the region is
Cancel the π's and move the fraction in front of the integral, and there's the formula:
Be Careful:
To find the area of a region in between two polar graphs, we can do it the hard way by adding areas directly. But we like to work smarter instead of harder. We are going find the larger area and subtract the smaller area.
Example 1
Find an integral expression for the area of the shaded region. | |
We can see from the picture that the limits of integration go from to . We slice the region into tiny slices and pretend the slice at angle θ is a slice of a perfect circle with radius r = sinθ + 1 for that particular value of θ. The area of the entire perfect circle is π(sinθ + 1)^{2} and the area of the tiny slice is a fraction of that: Adding up the areas of the tiny slices and letting the number of slices approach infinity, we see that the area of the shaded region is | |
Example 2
Find an integral expression for the area of the region enclosed by the graph of the polar function r = sin θ. | |
We have to be careful with the domain of θ. Since the sine function is periodic with period 2π, we might reasonably try the domain 0 ≤ θ ≤ 2π. This does seem to give us the right picture: However, this is a bigger interval than we need. If we use 0 ≤ θ ≤ π we get the same picture. If we try any smaller domain we don't get the whole picture. This means the limits of integration are α = 0 and β = π. To apply the formula we plug in those values for α and β and write sin θ in place of f(θ): That really is all there is to it. For these exercises, we should be aware that there may be more than one right answer. For example, the area of the region enclosed by the graph of r = sin θ is also given by If you get a different integral than the one given in the answer, use your calculator to evaluate your integral and also our integral. It's like choosing between one pizza topping or another: as long as you are satisfied, it's alright. For these problems, if both answers come out the same, you just found a different correct answer. | |
Example 3
Find an integral expression for the area of the region in between the graphs r = 2cos θ and r = cos θ. | |
We want to find the area of the region: Let's find the area inside the graph r = 2cos θ and subtract the area inside the graph r = cos θ. The area inside r = 2cos θ is and the area inside r = cos θ is So the area of the region in between the two graphs is Since the limits of integration on the two integrals are the same, we can combine them into the single integral We could also find this area one slice at a time. That is, we could find the area of the region between the graphs r = 2cos θ and r = cos θ by slicing the larger region into pizza slices, figuring out the area of the "crust" on each slice, and adding those areas up. The area of the full slice is and the area of the juicy center part is , so the area of the crust is This simplifies to When we add up all all the crust areas and let the number of pieces approach ∞, we get the integral To generalize with a nice, neat, pizza-making formula, when we have a graph r_{outer} and a graph r_{inner}, the area in between the graphs for α ≤ θ ≤ β is Be Careful: for the area between the graphs r_{outer} and r_{inner}. This may look less complicated, but it's wrong. The two radii must be squared separately and then subtracted. | |
Exercise 1
Write an integral expression for the area for ...
the shaded region
Answer
The function we need for the integrand is
f(θ) = sin θ.
We need to find the limits that make sense for θ in this region. The region starts at and ends at the smallest angle β in the second quadrant with a cosine of . That would be
We put α, β, and f(θ) into the formula to get an integral expression for hte area of the shaded region:
Exercise 2
Write an integral expression for the area for ...
the region below the x-axis and above the graph of the polar function r = sin θ – 1
Answer
If we graph r = sin θ – 1 for 0 ≤ θ ≤ π, we get the boundary of the region we're looking for, traced from left to right:
This means we want α = 0 and β = π. The area of the region is
Exercise 3
Write an integral expression for the area for ...
one petal of the graph of the polar function r = sin(3θ)
Answer
By playing with the graphs on a calculator, we find the values of θ that go with each petal. The area of the shaded petal is
The integrals
and
are also correct, as these give the areas of the other petals. Each petal has the same area as the other two petals.
Exercise 4
Write an integral expression for the area for ...
one petal of the graph of the polar function r = cos(2θ)
Answer
The graph r = cos(2θ) looks like this:
We play with the graph and find the values of θ that correspond to each petal. Using this information to determine the limits of integration, any of the following integrals would be correct:
Exercise 5
Write an integral expression for the area for ...
\item the intersection of the regions enclosed by r = sin θ and r = cos θ
Answer
This is the region whose area we're looking for:
If we look closely, we see that this region consists of 2 parts. One part is bounded just by the graph of cos θ, and the other part is bounded just by the graph of sin θ.
The part that's bounded by sin θ goes from θ = 0 to , and the part that's bounded by cos θ goes from to .
The area of the part bounded by sin θ is
and the area of the part bounded by cos θ is
Adding these, we find that the area of the region is
Exercise 6
Write an integral expression for the area
between the graphs r = sin θ and r = 2sin θ
Answer
We want the area of this region:
The angle θ goes from 0 to π as we go around this region once. The outer radius is r = 2sinθ and the inner radius is r = sin θ, so the area between the graphs is
Exercise 7
Write an integral expression for the area
of the shaded region
Answer
This region goes from θ = α to θ = β where these are the angles at which and the unit circle intersect.
Since r = 1 on the unit circle, we need
Solving, we find
We have and . The outer radius is and the inner radius is 1,
so the area of the region is