Video: Finding the Area of a Region in Polar Coordinates

Find the area of the region below the polar axis and enclosed by π‘Ÿ = 2 βˆ’ cos πœƒ.

05:08

Video Transcript

Find the area of the region below the polar axis and enclosed by π‘Ÿ is equal to two minus the cos of πœƒ.

We’re given a polar curve, and we’re asked to determine the area of the region which is below the polar axis and enclosed by our polar curve. To do this, the first thing we’re going to want to do is get a picture of our region. And to do this, we’re going to need to sketch our polar curve. There’s a few different ways we could do this. We could substitute values of πœƒ into our polar curve and then plot these onto a graph. However, the easiest way is to just use a graphing calculator. Using either method, we get a sketch which will look like the following.

Remember, though, we’re not just finding the area of the region bounded by this polar curve, but it also needs to be below the polar axis. In other words, our region is everything underneath the axis inside of our curve. Now, to find this area, we need to recall the following formula for finding the area bounded by the rays πœƒ one and πœƒ two and the polar curve defined by π‘Ÿ. It’s given by the integral from πœƒ one to πœƒ two of one-half π‘Ÿ squared dπœƒ. In our case, we’re going to want to find the area bounded by the polar curve given to us in the question. So π‘Ÿ is going to be equal to two minus the cos of πœƒ.

Now, we just need to determine suitable values for πœƒ one and πœƒ two. And there’s a few different ways of doing this. One way of doing this is to substitute some values of πœƒ into our polar curve. For example, if we substitute a few values around zero, we would see that our curve is traced counterclockwise. This is important to notice because when we choose our values of πœƒ one and πœƒ two, we need to make sure we take the correct half of our region. Now, by using the fact that our curve is traced counterclockwise, we can see the point between the second and third quadrant will be our value of πœƒ one. This is where our curve starts. And the point between the fourth and first quadrant will be where our curve ends, the value of πœƒ two.

We might be tempted to directly call these values πœ‹ and two πœ‹. However, we need to be careful. We don’t know yet if the value of π‘Ÿ is negative. However, we can just check both of these directly. We’ll substitute πœƒ is equal to πœ‹ and πœƒ is equal to two πœ‹ into our polar curve, giving us π‘Ÿ is equal to three and π‘Ÿ is equal to one, respectively. So these two values for πœƒ one and πœƒ two work. And remember, we’re choosing πœƒ two to be two πœ‹ and not zero because it’s the upper bound for our values of πœƒ.

So we’re now ready to determine the area of the region below the polar axis enclosed by our polar curve. It’s equal to the integral from πœ‹ to two πœ‹ of one-half times two minus the cos of πœƒ all squared with respect to πœƒ. The first thing we’ll do is take the constant factor of one-half outside of our integral. Then, we’ll distribute the square over our parentheses. This gives us one-half times the integral from πœ‹ to two πœ‹ of four minus four cos of πœƒ plus the cos squared of πœƒ with respect to πœƒ. And we can see that our integrand contains the cos squared of πœƒ.

There’s a few different ways of evaluating the integral of this. The easiest way is to use the double-angle formula for cosine. We recall the cos of two πœƒ is equivalent to two cos squared of πœƒ minus one. We want to rearrange this equivalence to make the cos squared of πœƒ the subject. We’ll add one to both sides of our equivalence and divide through by two. This gives us the cos of two πœƒ plus one all over two is equivalent to the cos squared of πœƒ. And we can see that this is a much-easier expression to integrate.

So we’ll substitute this expression into our integrand. This gives us the following expression which we need to integrate. And there’s one more piece of simplification we can do before we evaluate this. We can simplify four plus one-half to give us nine over two. And we can now just evaluate this integral term by term. First, the integral of the constant nine over two with respect to πœƒ is equal to nine πœƒ over two. Next, we know the integral of negative cos of πœƒ with respect to πœƒ is negative sin of πœƒ. So the integral of negative four cos of πœƒ with respect to πœƒ will be negative four sin of πœƒ.

And we can say something similar for our third and final term. When we integrate the cos of two πœƒ, we’ll get the sin of two πœƒ over two, so our third term integrates to give us the sin of two πœƒ over four. And this is a definite integral, so we don’t add our constant of integration, and we need to evaluate this at the limits of integration. Now, all that’s left to do is evaluate this at the limits of integration. We’ll start when πœƒ is equal to two πœ‹. Now we could just substitute this directly into our expression. However, we see the sin of two πœ‹ and the sin of four πœ‹ is equal to zero.

So if the second and third term evaluate to give us zero, then when we substitute two πœ‹ onto our antiderivative, we just get nine times two πœ‹ over two. And if we cancel the shared factor of two, we see this is just equal to nine πœ‹. Now we need to subtract our antiderivative evaluated at πœƒ is equal to πœ‹. However, we get a similar story. The sin of πœ‹ and the sin of two πœ‹ are equal to zero. So our only nonzero term will be nine πœ‹ by two. So we need to subtract this. We get one-half times nine πœ‹ minus nine πœ‹ by two. And if we evaluate this, we get nine over four times πœ‹, which is our final answer.

Therefore, we were able to show the area of the region below the polar axis enclosed by the polar curve π‘Ÿ is equal to two minus the cos of πœƒ is equal to nine over four times πœ‹.

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