Video: Finding the Area of a Region Bounded by a Given Curve

Find the area of the region bounded by the polar curve π‘Ÿ = 1 βˆ’ sin πœƒ.

06:18

Video Transcript

Find the area of the region bounded by the polar curve π‘Ÿ is equal to one minus the sin of πœƒ.

In this question, we’re given a curve defined by a polar equation. And we need to find the area of the region bounded by this polar curve. The first thing we’re going to want to do is determine the region bounded by our polar curve. We can do this by using a sketch. We could do this by using a graphing calculator or by substituting different angles of πœƒ into one minus the sin of πœƒ. We’ll use this method. We’ll start with πœƒ is equal to zero. We substitute this into one minus the sin of πœƒ, giving us one minus the sin of zero. And we know the sin of zero is equal to zero, so this simplifies to give us one. We now need to plot this on our polar diagram.

Remember, πœƒ is the angle our point will make with the positive π‘₯-axis and π‘Ÿ is the distance from the origin. The next value of πœƒ we will try is πœ‹ by four. Substituting this then to three decimal places, we get 0.293. We then need to plot this on our diagram. Next, we’ll try πœƒ is πœ‹ by two. We know the sin of πœ‹ by two is equal to one. So, one minus the sin of πœ‹ by two is equal to zero. Since the value of π‘Ÿ is equal to zero, this point will be at the origin. And we can do exactly the same with the following values of πœƒ. This gives us the following values for π‘Ÿ. And just like we did with the previous points, we can then plot these onto our polar diagram.

However, we’ll check one more value of πœƒ: when πœƒ is equal to two πœ‹. We’ll substitute this into our curve, giving us one minus the sin of two πœ‹ which is equal to one. This means we need to plot the point two πœ‹, one onto our polar diagram. However, we can see we’ve already plotted this point. It’s the same as the point when πœƒ is equal to zero. And this is because the sin of πœƒ is periodic about two πœ‹. We can then fill in the rest of our curve, and we can check this by using our graphing calculator. And before we find the area of our region bounded by the polar curve, we need to notice the following.

First, our curve started when πœƒ was equal to zero. And when πœƒ was equal to two πœ‹, we had made one full loop. This means our values of πœƒ will vary from zero to two πœ‹. Next, we can see from our sketch there are no inner loops. And this is useful. We can also see this from the definition of our curve. We see that π‘Ÿ must be greater than or equal to zero. And this is because the sin of πœƒ is always between positive and negative one, so the lowest our value of π‘Ÿ can be is zero. We’re now ready to start finding the area of the region bounded by our polar curve. We just need to recall the following.

We can find the area of a region bounded by a polar curve π‘Ÿ between the rays πœƒ one and πœƒ two by calculating the integral from πœƒ one to πœƒ two of one-half times π‘Ÿ squared with respect to πœƒ. And we’ve already shown our values of πœƒ will range from zero to two πœ‹, so we’ll set πœƒ one equal to zero and πœƒ two equal to two πœ‹. And in the question, we’re told that π‘Ÿ is equal to one minus the sin of πœƒ. Therefore, by using this value of πœƒ one, πœƒ two and our expression for π‘Ÿ, we can find the area of the region bounded by our polar curve. It’s equal to the integral from zero to two πœ‹ of one-half times one minus the sin of πœƒ all squared with respect to πœƒ.

So all we need to do is evaluate this integral. We’ll start by distributing the square over our parentheses. Doing this gives us the integral from zero to two πœ‹ of one-half times one minus two sin of πœƒ plus the sin squared of πœƒ with respect to πœƒ. Next, we’ll take the constant factor of one-half outside of our integral. Doing this gives us the following expression. And there’s several different methods we could use to evaluate this integral. However, we can see in our integrand we have the sin squared of πœƒ, and there’s no easy way to integrate the sin squared of πœƒ. So we’ll do this by using the double-angle formula for cosine.

We’ll start by recalling the following version of the double-angle formula for cosine. The cos of two πœƒ is equivalent to one minus two times the sin squared of πœƒ. We want to rearrange this to make sin squared of πœƒ the subject. To do this, we’ll subtract one from both sides of our equivalence and then divide through by negative two. We’ll then simplify. This gives us the sin squared of πœƒ is equivalent to one minus the cos of two πœƒ all divided by two. So instead of integrating the sin squared of πœƒ, we can instead integrate one minus the cos of two πœƒ all divided by two. And we can see this is much easier to integrate, so we’ll use this to rewrite our integrand.

So by using this, we’ve rewritten our integral as one-half times the integral from zero to two πœ‹ of one minus two sin of πœƒ plus one minus the cos of two πœƒ all over two with respect to πœƒ. And before we evaluate this integral, we’ll simplify this slightly. We have one plus one-half which we know is equal to three over two. So now we’ve rewritten our integral in the following form, and we can see we can evaluate this term by term. Let’s start with our first term. We can see that three over two is a constant, so its integral with respect to πœƒ will be three πœƒ over two. For the integral of our second term, we know the integral of negative the sin of πœƒ is equal to the cos of πœƒ, so the integral of negative two sin πœƒ with respect to πœƒ is two cos of πœƒ.

Finally, we know the integral of the cos of two πœƒ with respect to πœƒ is the sin of two πœƒ divided by two. So we can simplify our third integral to give us negative the sin of two πœƒ over four. And we don’t add a constant of integration because this is a definite integral. And we need to remember to add our limits of integration. All that’s left to do now is evaluate this at the limits of integration. We’ll start with our upper limits of integration when πœƒ is equal to two πœ‹, so we need to substitute πœƒ is equal to two πœ‹ into our antiderivative. This gives us the following expression.

And we can then simplify this expression. First, two divided by two is equal to one. Next, the cos of two πœ‹ is equal to one. And in our third term, we have two times two πœ‹ which is equal to four πœ‹. But then we know the sin of four πœ‹ is just equal to zero, so our third term is just equal to zero. So substituting two πœ‹ into our antiderivative gives us three πœ‹ plus two. We now need to subtract zero substituted into our antiderivative. We could do this in exactly the same way. However, we notice if we substitute πœƒ is equal to zero into our antiderivative, our first and third term will evaluate to give us zero.

So in fact, we only need to subtract the second term in our antiderivative evaluated at zero. In other words, we’re subtracting two times the cos of zero. This gives us one-half times three πœ‹ plus two minus two times the cos of zero. And of course, we can simplify this. First, the cos of zero is equal to one, but then two minus two is equal to zero. So this entire expression simplifies to give us three πœ‹ by two which is our final answer. Therefore, we were able to show the area of the region bounded by the polar curve π‘Ÿ is equal to one minus the sin of πœƒ is equal to three πœ‹ by two.

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