For the following exercises, find the area of the described region. Common interior of r=4 sin(2

For the following exercises, find the area of the described...

Key Concepts

Polar Coordinates

This concept involves representing points in a plane by their distance from a reference point (the pole) and the angle from a reference direction. It's fundamental to understanding how curves and regions are expressed in forms like r = f(?), which are central in many calculus problems involving circular and radial symmetry.

Area Calculation in Polar Coordinates

This key area of study explains how to compute areas of regions defined by polar curves using the integral formula ½?(r(?))^2 d?. Understanding this method is crucial when dealing with regions bounded by curves represented in polar form.

Intersection of Polar Curves

Determining the intersection points of given polar curves is essential for setting the correct limits of integration. This concept involves solving the equation f(?) = g(?) to find the angles at which two curves meet, thereby defining the boundaries of the area to be computed.

Symmetry in Polar Graphs

Recognizing symmetry can significantly simplify computations by reducing the interval of integration. Many polar curves exhibit symmetry (about the polar axis, the line ? = ?/2, or the pole), which can be exploited to compute areas more efficiently by evaluating a portion of the region and then multiplying to obtain the total area.

Transcript

00:01 So here we are supposed to find the region which is enclosed or which is a common for both this curves.

00:05 So the curves that are given to us are r is equal to we have 4 sine 2 theta.

00:13 So this is in the form of a rose which has you can see in the diagram which has four petals.

00:19 Okay.

00:20 And the next is r is equal to 2.

00:24 So this you can see i have drawn it is a circle with the center as 0 .0 and the 3.

00:32 Radius is equal to 2.

00:33 So now i'm representing this part, the red part here is the rows.

00:37 So the equation for this is r equal to 4, sine, 2, theta.

00:42 Okay, and the circle over here is representing the curve are equal to 2.

00:49 Now, they have told us to consider the common areas for both the curves.

00:53 So common areas you can see are actually in four parts.

00:55 So this is the first part, then this is the second part, third part and four part.

01:00 But since they are symmetrical, i will solve only for the quarter part which is of the first quadrant and i will multiply the answer by four to get the total answer so then to solve for the first part you can see that the common area here is the path of this one okay then we have this path and then the next is this part so if i consider them into three areas if this is going to be a1 this will be a2 and this path will be a3 all right so now uh to find the angles or the value of theta from where it is uh ranging we need to first find the intersecting points of the angles at these intersecting points okay so then if i have to highlight those angles here then i'm going to extend this line and even extend this line and we know that theta over here is zero and theta and this line is pi by 2 all right so we're supposed to find theta here okay and theta here so to find that we know that these are the two intersecting points so to find the intersecting points and the value of theta there we are going to equate both the values so we have four sine 2 theta is equal to 2 so i therefore get sine 2 theta is equal to half and therefore we have 2 theta is equal to sine inverse of half so that is going going to give you pi by six and therefore i'm getting the value okay this was for the first quadrant and this will again um i mean there are two intersecting points so the first value we have is pi by six and the other we have is five pi by 12 okay oh sorry five pi by six yeah this is five by six because later on months we divide by two this will change to pi by 12 and five by 12 okay so this value here is pi by 12 and this value here is 5 pi by 12 so now to remember one important part here is that you can see in the diagram that for a 1 we are going from 0 the angle is going from 0 to pi by 2 then for a 2 we are going from pi by 2 sorry pi by 12 so the first one is from 0 to pi by 12 then we are going from pi by 12 to 5 pi by 12 and then from 5 pi pi by 12 we are coming back to pi by 2 so the area is divided into three parts here.

03:52 So to start with, i'm going to write down the formula for the area.

03:58 Now, the formula for area in general is given by half integral from alpha to beta.

04:05 We have f of theta, which is r, the whole square d theta, right? so since our area is divided into three parts, i'm going to write down the total area here, a.

04:16 So that is equal to we have half...

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