WEBVTT
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Area Bounded by Polar Curves
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In this video, we will learn how to calculate the area of a region enclosed by one or more polar curves.
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Weβll be looking at a variety of examples of how we can find integrals to find areas of this form.
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Letβs now consider the following polar curve.
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And we can call this curve π or π of π.
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Letβs suppose that we were asked to find the area bounded by this curve, π one and the π two.
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Now, typically, when weβre finding an area in Cartesian coordinates, the area we will be finding here would be the area between the curve π, the horizontal π₯-axis, and the π₯-values of π one and π two.
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So thatβs this area here.
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However, this is not a Cartesian graph.
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This is a polar graph.
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And when we say weβre finding the area between π one and π two, this, in fact, means the area enclosed by the lines π is equal to π one, π is equal to π two, and π is equal to π of π.
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Which gives us this shaded area in blue here.
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Which, as we can see, is not equal to the area we would have found using the Cartesian method, which is in yellow.
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Therefore, we have a new formula for finding this area.
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This formula tells us that the area is equal to the integral from π one to π two of one-half π squared dπ.
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Letβs now look at an example of how this formula can be used.
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Find the area of the region bounded by the polar curve π is equal to two cos two π between π is equal to π by two and π is equal to three π by two.
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Now, we have a formula for finding area bounded by polar curves.
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And this formula tells us that the area is equal to the integral from π one to π two of one-half π squared dπ.
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Now, weβve been given π in the question, and itβs equal to two cos of two π.
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And weβve been asked to find the area between π is equal to π by two and π is equal to three π by two.
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Since three π by two is greater than π by two, we can say that π by two must be equal to π one.
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And three π by two must be equal to π two.
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We can substitute these values of π one, π two, and π into our equation for the area.
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We find that the area is equal to the integral from π by two to three π by two of one-half of two cos of two π squared dπ.
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We can start by expanding the square term.
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We end up with the integral from π by two to three π by two of one-half times four cos squared of two π dπ.
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And we can cancel the half with one of the twos from the four, which gives us this.
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Now, we cannot directly integrate a cos squared term like this on its own.
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However, there is a way to get rid of the square.
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We need to use the double angle formula for cos.
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This formula tells us that cos of two π is equal to two cos squared π minus one.
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We can rearrange this to make cos squared π the subject, which gives us that cos squared π is equal to cos of two π plus one all over two.
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Now, weβre nearly ready to substitute this formula in.
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However, in our integrand, we, in fact, have cos squared of two π.
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And in our formula, we only have cos squared of π.
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This can be solved by simply multiplying each π in our formula by two.
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Which gives us that cos squared of two π is equal to cos of four π plus one all over too.
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And we can substitute this back into our integral to obtain the integral from π by two to three π by two of two multiplied by cos of four π plus one all over two dπ.
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And now, we can cancel out the factor of two with the two in the denominator.
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And now, weβre ready to integrate our function.
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Our first term, which is cos of four π, is a function within a function.
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The innermost function is four π.
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So we start by differentiating four π with respect to π to give us four.
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Since four is a constant, weβre able to do this integration.
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However, we mustnβt forget to divide by this four.
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And so, we start off with one over four.
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Then, we integrate the cos.
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And so, for our first term, we obtain one over four sin of four π.
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And our second term is simply one.
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Integrating one just gives us π.
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And we mustnβt forget that weβre integrating between π by two and three π by two.
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So these are the bounds which we need to substitute in.
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Substituting in three π by two, we obtain one-quarter sin of six π plus three π by two.
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And next, we substitute in π by two, but we mustnβt forget to subtract this.
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Since π by two is our lower bound.
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And so, we obtain minus one over four sin of two π plus π by two.
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Now, we know that sin of two π is equal to zero.
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And we also know that sin is a periodic function with a periodicity of two π.
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Since six π is a multiple of two π, this means that sin of six π will also be equal to zero.
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Therefore, the two terms which weβre left with is three π by two minus π by two.
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And itβs from here that we reach our solution.
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Which is that the area of the region bounded by the polar curve π is equal to two cos of two π between π is equal to π by two and π is equal to three π by two is equal to π.
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In some questions, we may not be given the values of π one and π two, for which we need to find the area between.
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Sometimes, the region of which we need to find the area will simply be described.
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We will see this in the next example.
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Find the area of the region enclosed by the inner loop of the polar curve π is equal to one plus two sin π.
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Letβs start by drawing a quick sketch of what this curve would look like.
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We could do this by using a graphing calculator, graphing software, or by plotting some points on the curve.
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Our curve would look something like this.
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We can see that the inner loop of our curve is this little loop at the bottom here.
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And so, this orange-shaded region is the area weβre trying to find.
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We do, in fact, have a formula for finding areas of regions enclosed by polar curves.
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This formula tells us that the area is equal to the integral from π one to π two of one-half π squared dπ.
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We have been given π in the question.
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And itβs equal to one plus two sin π.
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We just need to find the values of π one and π two.
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Now, the values of π one and π two will occur at the end points of their closed inner loop.
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Since this is a loop, this will happen when the curve crosses over itself.
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Therefore, this is at this blue point here.
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We can see from our sketch of the curve that this occurs when π is equal to zero, since this point is on the origin.
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And so, we can find π one and π two by solving π is equal to zero.
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When π is equal to zero, one plus two sin π is equal to zero too.
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We can rearrange this to find that sin π is equal to negative one-half.
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We need to find solutions here within the range, where π is between zero and two π.
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We can use a graph of sin π to help us find our solutions.
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We draw in a line of negative one-half.
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Our calculator will give us a solution of negative π by six.
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However, weβre looking for solutions between zero and two π.
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Since sin is periodic in two π, we can see from our sketch that one of the solutions will be π by six less than two π, which is 11π by six.
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And the other solution will be π by six more than π, which is simply seven π by six.
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Therefore, weβve now found our two values for π one and π two.
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We have that π one is equal to seven π by six and π two is equal to 11π by six.
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Weβre now ready to use our formula for finding the area.
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We have that the area is equal to the integral from seven π by six to 11π by six of one-half multiplied by one plus two sin π squared dπ.
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We can distribute the square and multiply through by one-half.
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Which gives the integral from seven π by six to 11π by six of one-half plus two sin π plus two sin squared π dπ.
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Now, we know how to integrate each of these terms apart from the sin squared term.
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However, we can rewrite the sin squared term using the double-angle formula for cos.
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We have that cos of two π is equal to one minus two sin squared π.
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This can be rearranged to make sin squared π the subject.
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Which gives that sin squared π is equal to one minus cos of two π all over two.
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And this can be substituted back into our integral.
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And we can cancel the factor of two with the factor of two in the denominator.
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Leaving us with the integral from seven π by six to 11π by six of one-half plus two sin π plus one minus cos of two π dπ.
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And we can group together the one and the one-half.
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So our integrand ends up looking like this.
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Which weβre now ready to integrate.
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Integrating the first term, three over two, we simply get three π over two.
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When we integrate the second term of two sin π, we get negative two cos π.
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And for our final term of negative cos two π, we get negative one-half sin of two π.
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And we mustnβt forget that this is between the bounds of seven π by six and 11π by six.
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Now, we start by substituting in 11π by six, giving us 11π by four minus two cos of 11π by six minus one-half sin of 11π by three.
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Next, we substitute in seven π by six.
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But you mustnβt forget to subtract this since seven π by six is the lower bound.
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Giving us negative seven π by four minus two cos of seven π by six minus one-half sin of seven π by three.
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Next, we use the fact that cos of 11π by six is equal to root three over two.
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Sin of 11π by three is equal to negative root three over two.
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Cos of seven π by six is equal to negative root three over two.
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And sin of seven π by three is equal to root three over two.
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We can now multiply through all of the terms.
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And we end up with 11π by four minus seven π by four minus root three plus root three over four minus root three plus root three over four.
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Combining these terms together, we reach our solution.
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Which is that the area of the region enclosed by the inner loop of π is equal to one plus two sin π is equal to π minus three root three over two.
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Next, we will learn how to find the area enclosed by two polar curves.
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Letβs consider the following scenario.
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Letβs say weβve been given two polar curves π one and π two and weβve been asked to find the region enclosed by π one and π two between π one and π two.
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So thatβs this shaded region here.
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We can find a formula for this area by applying the formulas which we already know for finding areas of regions bound by polar curves.
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We know that we can find the area enclosed by π two, π one, and π two, which is this area here.
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And itβs equal to the integral from π one to π two of one-half π two squared dπ.
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We also know how to find the area enclosed by π one, π one, and π two, which is this area here.
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We know that itβs equal to the integral from π one to π two of one-half π one squared dπ.
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Now, we can see that this yellow area, which weβre trying to find, must be equal to the area encased by π one between π one and π two minus the area encased by π two between π one and π two.
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So thatβs the integral between π one and π two of one-half π one squared dπ minus that integral between π one and π two of one-half π two squared dπ.
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Using integral properties, weβre able to combine these integrals since they have the same bounds, to give the integral between π one and π two of one-half π one squared minus one-half π two squared dπ.
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We can factor out the half to form our final formula.
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Which is that the area between two curves π one and π two between π one and π two is equal to the integral between π one and π two of one-half π one squared minus π two squared dπ.
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Letβs note that this only works when π one is greater than or equal to π two for π-values between π one and π two.
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Now that we formed this new formula, letβs look at an example of how it works.
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Find the area of the region that lies inside the polar curve π squared is equal to eight cos of two π but outside the polar curve π is equal to two.
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In order to find the regions that lie outside π is equal to two but inside π squared is equal to eight cos of two π, weβll need to find the points of intersections of the two curves.
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Therefore, we can start by doing this.
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We need to solve π squared is equal to eight cos of two π and π is equal to two, simultaneously.
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We can substitute π equals two into the first equation.
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This will give us that four is equal to eight cos of two π.
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We end up with cos of two π is equal to one-half.
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Now, weβre looking for the solutions for which π is between zero and two π.
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Therefore, these will be the solutions for which two π is between zero and four π.
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These solutions within this range are that two π is equal to π by three, five π by three, seven π by three, or 11π by three.
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And so, we find that our points of intersection are that π is equal to π by six, five π by six, seven π by six, and 11π by six.
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Letβs now draw a sketch of π is equal to two.
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This will simply be a circle with a radius of two.
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Next, we can mark on our points where the two curves will intersect.
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The first point is at π by three.
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The second is at five π by three.
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The third is at seven π by three.
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And the fourth intersection point is at 11π by three.
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Now, we can attempt to sketch the curve of π squared is equal to eight cos of two π.
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We can do this by first rewriting the curve as π is equal to the square root of eight cos of two π.
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Next, we can find some points on the curve by inputting some values of π.
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We have that at π is equal to zero, cos of zero is equal to one.
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Therefore, π is equal to the square root of eight, which is also equal to two root two.
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At π is equal to π by two, we have cos of two π is equal to cos of π, and cos of π is equal to zero.
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Therefore, at π is equal to π by two, π is equal to zero.
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Continuing this on, at π is equal to π, we have that π is equal to two root two.
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And at π is equal to three π by two, π is equal to zero.
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And we can add these points to our graph.
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We now have seven points of our curve drawn on our graph.
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And so, we can draw a rough sketch for what it should look like.
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As we can see, it sort of looks like a figure of eight.
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Now, this sketch really helps us to find the regions of which weβre trying to find their areas.
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So itβs the regions which lie outside the curve π is equal to two but inside our curve of π squared is equal to eight cos of two π.
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There are, in fact, two regions which weβre interested in, which is these two regions here.
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We have a formula for finding the area of regions between two polar curves.
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And this formula tells us that the area is equal to the integral between π one and π two of one-half π one squared minus π two squared dπ.
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For the regions which weβre concerned with, π squared is equal to eight cos of two π is greater than or equal to π is equal to two.
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Therefore, π one squared is equal to eight cos of two π, and π two is equal to two.
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We can start by considering the area on the left of our diagram.
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This will be equal to the integral from five π by six to seven π by six of one-half of eight cos of two π minus four dπ.
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Next, letβs consider the area of the region on the right of our graph.
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This region is between the angles of 11π by six and π by six.
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However, if we were to integrate between 11π by six and π by six, we would be jumping from an angle of two π to an angle of zero.
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Which would occur when we cross the horizontal access.
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And, of course, we cannot do this.
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We can instead change the angle of 11π by six to negative π by six.
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Since on our graph, negative π by six is equal to 11π by six.
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We obtained that the area on the right is equal to the integral from negative π by six to π by six of one-half of eight cos of two π minus four dπ.
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We can simplify both of these integrands.
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And now, weβre ready to integrate.
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We have that the integral of four cos of two π minus two is two sin of two π minus two π.
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Next, we substitute in our upper and lower bounds.
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Now, we can use that sin of seven π by three and sin of π by three are both root three over two.
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And that sin of five π by three and sin of negative π by three are both negative root three over two.
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Next, we can expand everything here.
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And then, for our final step, we just simplify.
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Which gives us a solution that the area of the region that lies inside π squared is equal to eight cos of two π but outside π is equal to two is four root three minus four π by three.
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We have now seen a variety of examples and techniques which we can use to find the areas bound by polar curves.
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Letβs cover some key points of the video.
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Key Points
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The area enclosed by the polar curve π is equal to π of π, π one, and π two is given by the integral from π one to π two of one-half π squared dπ.
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The area enclosed by two polar curves π one which is equal to π of π and π two which is equal to π of π, π one, and π two.
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Where π one is greater than or equal to π two for π between π one and π two.
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Is given by the integral from π one to π two of one-half of π one squared minus π two squared dπ.