Video: Finding the Area Enclosed by a Parametrically Defined Curve

Consider the curve defined by the parametric equations π‘₯ = 2 cos 𝑑 and 𝑦 = 3 sin 𝑑. Find ∫ βˆ’6 sinΒ² 𝑑 d𝑑. Find the area under the curve when 0 ≀ 𝑑 ≀ πœ‹. Now, by taking 0 ≀ 𝑑 ≀ 2πœ‹, find the total area inside the curve.

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Video Transcript

Consider the curve defined by the parametric equations π‘₯ equals two cos 𝑑 and 𝑦 equals three sin 𝑑. Find the indefinite integral of negative six sin squared 𝑑 with respect to 𝑑. Find the area under the curve when 𝑑 is greater than or equal to zero and less than or equal to πœ‹. Now, by taking 𝑑 is greater than or equal to zero and less than or equal to two πœ‹, find the total area inside the curve.

Now, here, we’ve been given the equation of a curve defined parametrically. And the first part of this question really doesn’t seem to have anything to do with the setup. We’re looking to integrate negative six sin squared 𝑑 with respect to 𝑑. But actually, these things are intrinsically linked, and we’ll see how in a moment.

But let’s first begin by evaluating that integral. We’re going to use a trigonometric identity to find this integral. We know that sin squared 𝑑 is equal to one minus cos two 𝑑 all divided by two. And so, actually, we can evaluate the indefinite integral of negative six times one minus cos two 𝑑 over two with respect to 𝑑.

Let’s begin by dividing through by a factor of two. And we also know that we can take any constant factors outside of our integral and focus on integrating the function itself. So we have negative three times the integral of one minus cos of two 𝑑 with respect to 𝑑.

Well, we can integrate this term by term. The integral of one with respect to 𝑑 is fairly straightforward. It’s simply 𝑑. We also know that the integral of cos of π‘Žπ‘₯ with respect to π‘₯, where π‘Ž is some real constant, is one over π‘Ž times sin of π‘Žπ‘₯ plus of course some constant of integration 𝑐. So the integral of negative cos of two 𝑑 is negative one over two times sin of two 𝑑. This is an indefinite integral. So we’ll add a constant of integration. Let’s call that 𝐴 for now.

Our final step is to distribute our parentheses. By multiplying each term by negative three, we get negative three 𝑑 plus three over two times sin of two 𝑑 plus 𝑐. Now we’ve changed that from 𝐴 to 𝑐 because technically our earlier constant has been multiplied by negative three. So it’s now a different value.

The second part of this question asks us to find the area under the curve when 𝑑 is greater than or equal to zero and less than or equal to πœ‹. Now rather than trying to define our curve using a Cartesian equation β€” i.e., 𝑦 is some function of π‘₯. We recall that the area of a curve defined parametrically by the functions π‘₯ of 𝑑 and 𝑦 of 𝑑 for values of 𝑑 between π‘Ž and 𝑏 is the definite integral between π‘Ž and 𝑏 of 𝑦 of 𝑑 times π‘₯ prime of 𝑑 with respect to 𝑑.

Now of course π‘₯ prime of 𝑑 is the derivative of π‘₯ with respect to 𝑑. So, we might write that as dπ‘₯ by d𝑑. And we recall that π‘₯ is two cos 𝑑 and that the derivative of cos of π‘₯ with respect to π‘₯ is negative sin π‘₯. And so dπ‘₯ by d𝑑 must be negative two sin 𝑑. And so, the area under the curve for 𝑑 is greater than or equal to zero and less than or equal to πœ‹ is the definite integral between zero and πœ‹ of three sin 𝑑 times that derivative, times negative two sin 𝑑 with respect to 𝑑. And if we simplify that integral, we get negative six sin squared 𝑑.

Now we integrated that in the first part of this question. Since we have a definite integral, the constants of integration end up canceling. So, actually, we just need to evaluate negative three 𝑑 plus three over two sin of two 𝑑 between πœ‹ and zero. That’s negative three πœ‹ plus three over two sin of two πœ‹ minus negative three times zero plus three over two times sin of two times zero.

Now, quite strangely, that integral gives us negative three πœ‹. But of course, we know we’re dealing with an area. Now it’s note though when we sketch this curve in the Cartesian plane, it lies below the π‘₯-axis. In fact, the opposite is true. The reason the integral yields a negative answer here is due to the fact that the curve is traced from right to left rather than left to right. And so, the area under our curve when 𝑑 is greater than or equal to zero and less than or equal to πœ‹ is three πœ‹ or three πœ‹ square units.

Now, the third part of this question asks us to take 𝑑 from greater than or equal to zero and less than or equal to two πœ‹. Now, if we were to sketch this curve out, we get this sort of oval shape as shown. We’re completing a full turn, and so there is no overlapping. The area is therefore the definite integral between zero and two πœ‹ of negative six sin squared 𝑑 with respect to 𝑑. This time, substituting zero and two πœ‹ into our earlier integral. And we get negative six πœ‹ plus three over two sin of four πœ‹ take away zero. Which this time gives us negative six πœ‹.

Now, once again, the negative is because the curve is being sketched out from right to left rather than left to right. And so, we say the total area inside the curve is six πœ‹ or six πœ‹ square units.

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