Which graph represents the polar curve r 4sin(θ)

Find the area of the region that lies inside the polar curve 𝑟 equals four sin 𝜃 but outside the polar curve 𝑟 equals two.

In order to answer the question, let’s sketch the two given polar curves. Let’s start by sketching the polar curve 𝑟 equals two, as it is slightly easier to sketch than the polar curve 𝑟 equals four sin 𝜃. The polar curve 𝑟 equals two consists of all points of the form 𝑟, 𝜃, where 𝑟, the distance from the origin 𝑂 to the point, is equal to two and 𝜃, the angle between the polar axis and the point, is allowed to take any possible value. We therefore deduce that the polar curve 𝑟 equals two represents the circle of radius two centered at the origin.

Let’s now sketch the polar curve 𝑟 equals four sin 𝜃. In order to do this, let’s first sketch the graph of 𝑟 equals four sin 𝜃 in Cartesian coordinates. This is just the graph of 𝑟 equals sin 𝜃 stretched vertically by a scale factor of four. Sketching the graph of 𝑟 equals four sin 𝜃 in Cartesian coordinates enables us to read at a glance the values of 𝑟 that correspond to the increasing values of 𝜃. We see that as 𝜃 increases from zero to 𝜋 over two, 𝑟, the distance from the origin, increases from zero to four. This means that the first quadrant of the polar graph of 𝑟 equals four sin 𝜃 looks like this.

We see that as 𝜃 increases from 𝜋 over two to 𝜋, 𝑟 decreases from four to zero. This means that the second quadrant of the polar graph of 𝑟 equals four sin 𝜃 looks like this. If we let 𝜃 increase beyond 𝜋 or decrease beyond zero, then we just retrace our path. Either by plotting more points and observing the symmetry. Or, by converting the polar equation 𝑟 equals four sin 𝜃 into its equivalent Cartesian equation by using 𝑦 equals 𝑟 sin 𝜃 and 𝑥 squared plus 𝑦 squared equals 𝑟 squared. And then, completing the square, we find that the polar curve 𝑟 equals four sin 𝜃 is a circle of diameter four — i.e., radius two — centered at the Cartesian coordinates zero, two.

Sketching the two polar curves 𝑟 equals two and 𝑟 equals four sin 𝜃 on the same graph, we obtain two overlapping circles of radius two, like so. Note that we could have also sketched these curves using a graphing calculator instead of doing so by hand. The question asks us to find the area of the region that lies inside the polar curve 𝑟 equals four sin 𝜃 but outside the polar curve 𝑟 equals two. This corresponds to the region shaded in green. Recall that the area of a polar curve 𝑟 equals 𝑓 of 𝜃 between 𝜃 equals 𝜃 one and 𝜃 equals 𝜃 two is the integral from 𝜃 one to 𝜃 two of a half times 𝑟 squared with respect to 𝜃.

If we let 𝜃 one and 𝜃 two be the angles associated to the points of intersection of the two curves 𝑟 equals four sin 𝜃 and 𝑟 equals two. Then we see that we can obtain the required area by computing the area between 𝜃 one and 𝜃 two of the polar curve 𝑟 equals four sin 𝜃. And subtracting from it the area between 𝜃 one and 𝜃 two of the polar curve 𝑟 equals two. Using the formula for the area of a polar curve, we can translate this as follows. The required area is equal to the integral from 𝜃 one to 𝜃 two of a half times four sin 𝜃 all squared with respect to 𝜃 minus the integral from 𝜃 one to 𝜃 two of a half times two squared with respect to 𝜃. Let’s work out 𝜃 one and 𝜃 two which are the 𝜃-values for the points of intersection of the two curves given to us in the question.

In order to do this, we will solve the equations 𝑟 equals four sin 𝜃 and 𝑟 equals two, simultaneously. The first step is to equate four sin 𝜃 and two. Dividing both sides of this equation by four and then simplifying the right-hand side, we obtain sin of 𝜃 equals a half. We obtain that 𝜃 is equal to the inverse sin of a half. Which we can compute to be 𝜋 over six using a scientific calculator or the illustrated special right-angled triangle. Since sin of 𝜃 is also equal to sin of 𝜋 minus 𝜃, we have that sin of 𝜋 by six is also equal to sin of 𝜋 minus 𝜋 by six, which is sin of five 𝜋 by six. So the inverse sin of a half also equals five 𝜋 by six.

We can check using the sin graph that 𝜋 by six and five 𝜋 by six are the only values for 𝜃 between zero and 𝜋 such that sin of 𝜃 equals a half. Therefore, the two polar curves given to us in the question intersect at 𝜃 one equals 𝜋 by six and 𝜃 two equals five 𝜋 by six. Let’s now evaluate the integrals to obtain the required area.

Using the property that the difference of integrals is the integral of the difference and the fact that we can pull constants outside of an integral, we obtain that the required area is equal to a half timesed by the integral from 𝜋 by six to five 𝜋 by six of four sin 𝜃 all squared minus two squared with respect to 𝜃. Distributing the brackets and the integrand, we obtain that the integrand is equal to 16 sin squared 𝜃 minus four. We can simplify the integral by taking a factor of four out, like so. Now, in order to integrate sin squared 𝜃, let’s substitute it with the equivalent expression a half times one minus cos two 𝜃. Simplifying, we obtain that the required area is equal to two times the integral from 𝜋 by six to five 𝜋 by six of one minus two cos two 𝜃 with respect to 𝜃.

Using the fact that the integral of a constant 𝑎 with respect to 𝜃 is 𝑎𝜃 and the fact that the integral of cos 𝑎𝜃 with respect to 𝜃, where 𝑎 is a constant, is one over 𝑎 sin 𝑎𝜃. We have that our required area is equal to two times 𝜃 minus two times a half times sin of two 𝜃 evaluated from 𝜋 by six to five 𝜋 by six. We can simplify this by cancelling out the two twos in the second term. Substituting in 𝜃 equals five 𝜋 by six and 𝜃 equals 𝜋 by six and taking the difference of the resulting expressions in that order, we obtain this expression. Simplifying, we obtain two times four 𝜋 by six minus sin of five 𝜋 by three plus sin of 𝜋 by three.

Using the special right-angled triangle illustrated earlier and the sine graph or using a scientific calculator, we can compute sin of five 𝜋 by three to be negative root three over two. We can compute sin of 𝜋 by three to be positive root three over two. Substituting these values in and then simplifying further, we obtain that the required area is equal to four 𝜋 over three plus two times the square root of three, which is our final answer.