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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 = sin θ + 1
for that particular value of θ. The area of the entire perfect circle is
π(sin θ + 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
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're satisfied, it's alright. For these problems, if both answers come out the same, you just found a different correct answer.
Find an integral expression for the area of the region in between the graphs r = 2cos θ and r = cos θ.
We're gunning for the area of this region here:
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 router and a graph rinner, the area in between the graphs for α ≤ θ ≤ β is
When working with two different radii, we don't want the formula below,
for the area between the graphs router and rinner.
This may look less complicated, but it's wrong. The two radii must be squared separately and then subtracted.