 Hello everyone, welcome to Centrum Academy YouTube channel where we bring for you every day a new question in order of problem-solving especially for the JEE main and advanced aspirants So guys and girls today we have a question which asks you to find the area in close in this curve I can see there's a curve which is given to you, which is not that familiar one, right? It is not something that we can easily relate to, right? So sketching this curve is going to be a tough ask So how do we solve these kind of questions? Let's try to understand So first of all a very careful observation here would show you that the terms that you have on the left side are Homogeneous in nature that means each of these terms are of degree 4 and The terms on the right side also are homogeneous. That means they are of the same degree and that degree is 2. Okay Now for such kind of questions we prefer using polar coordinates Yes polar coordinates could be a good help for us in such kind of questions So let's substitute x as r cos theta and let's substitute y as r sin theta Okay, so when you substitute this in this given Cartesian form, let's see what happens to the entire expression So x to the power 4 will be r to the power 4 so we can safely pull out r to the power 4 from each of these terms So you'll end up getting cos to the power 4 theta minus cos square theta sin square theta plus sin to the power 4 theta Okay, and on the right hand side you will have r square Times cos square plus sin square which is going to be a 1 so that leaves you with just an r square Right, so far so good. No issues. Okay, so we can drop off one of the r squares from both the sides Okay, so we can cancel off r squares from both the sides and not only that we can club these two terms we can club these two terms as cos square plus sin square the whole square minus 2 cos square sin square theta, right And of course we have an additional term of minus cos square sin square theta So in light of that this given expression boils down to r squared pi times 1 minus 3 sin square cos square theta equal to 1 Okay, in short you have r squared as 1 by pi times 1 minus 3 sin square theta cos square theta, okay So multiply the numerator and denominator with a 4 so that we can convert we can convert 4 sin square cos square theta as sin of 2 theta square Yes, so ladies and gentlemen, I'm now going to convert the denominator as 4 minus 3 sin square 2 theta Okay So now how do we plot this polar curve? So now that we need to plot it dear students. We need to keep certain things in our mind R represents the distance of a point from the origin Right, R represents the distance of a point on the curve from the origin or the pole, right? And theta basically here is the angle made by The line joining the point Joining the point With the x-axis Get it with the x-axis. Okay, so first of all we need to see when theta is 0 Okay, what is the value that R takes here? So when theta is 0 our square value is going to be 1 upon pi Okay, that means R is 1 upon root pi This value is very close to point five six Okay All right, now this is the value that is going to repeat Again when theta becomes a pi by 2 Correct, so this value this value is repeated for theta equal to 0 theta equal to pi by 2 theta is equal to pi theta equal to 3 pi by 2 and theta equal to 2 pi finally So can I say the graph is going to cut the x-axis the positive x-axis the negative x-axis The positive y-axis and the negative y-axis at the very same distance from the origin. Okay Now at the same time we also need to figure out what is the maximum value that R can take Okay, now all of you please understand the maximum value that R can take Is when this value becomes a 1 because then the difference of 4 minus 3 sin square 2 theta would be the least correct, so so R becomes maximum when sin of 2 theta square becomes a 1 and that will happen when 2 theta takes values like Pi by 2 that means theta could be pi by 4 It will also be maximum when 2 theta is 3 pi by 2 correct So when 2 theta is 3 pi by 2 my theta value will be 3 pi by 4 Okay, and theta is equal to 5 pi by 2 which means theta can be 5 pi by 4 And not only that 2 theta could be 9 pi by 2 which means theta is equal to 9 pi by 4, right? All right, so having figured this out Let's also figure out what could be this maximum value of R What could be this maximum value of R? So as I already said for you to have a maximum value of R sin square 2 theta should be a 1 Right, so if that is a 1 you'll end up getting 4 divided by pi as your R square So R max square is 4 by pi That means R max is going to be 2 by under root pi. So ladies and gentlemen, let's see how much thus 2 by root pi come out to be That comes out roughly to be 1.13 ish. Okay, so let's look into the graph of this function All right, so when you're drawing the graph of this function We need to be very careful that this graph is going to Basically show the same values of R for a change in the value of theta by by pi by 2 So let's say we are at this point when theta is 0 and our value is roughly 0.56. Okay. Now this value grows and becomes maximum of around So our value here becomes a maximum of around 1.13 ish at an angle of pi by 4 Okay, and slowly the value dies down and comes to the same value of 0.56 as it was when it started Okay, and now again it regains the value and goes to a maximum value of 1.13 at 3 pi by 4 Comes down back again to the same value point five six when theta is equal to pi Then goes again and takes the value of 1.13 at theta equal to 5 5 by 4 again returns back 2.56 when theta value is 3 pi by 2 again goes and becomes max at 1.13 when theta is 9 pi by 4 and finally returns to the same value when theta is equal to 2 pi Okay, so do you all what we see is this part of the curve this part of the curve if you see this the one which I am shading with yellow color This part of the curve is going to be repeated eight times This is going to be repeated eight times in this entire graph Right. So what I'm going to do is I'm going to use my formula for area of the curve in polar coordinate system, which is half r square d theta Okay, so let's use this fact half r square d theta Okay, so in our case my area will be obtained to be eight times half R square now, what's my r square? Let's go back to this figure. This is my r square my dear friends This is my r square. Okay, so let me put this expression in place of r square so that comes out to be for 4 by 4 minus 3 sine square 2 theta Okay into d theta into d theta Okay, so I'm going to integrate only from 0 to pi by 4 because I'm decided to take 8 times this area so you have to start from theta equal to 0 and Go all the way go all the way till theta equal to pi by 4, right? So let's try to put that and now let's try to simplify this So this is going to be this is going to be So I sorry I forgot that pi there. Yes, so it's going to be 16 upon pi 0 to pi by 4 D theta By 4 minus 3 sine square 2 theta Yes Now this is one of the integrals that we have already learned how to do it So let's divide the numerator and denominator by Cos square 2 theta So when you divide your numerator and denominator by cos square 2 theta you end up getting secant square 2 theta on the top You end up getting 4 secant square theta in the denominator, which you can also write as 1 plus tan square 2 theta and Minus 3 tan square 2 theta, right? So this becomes 16 by pi 0 to pi by 4 secant square 2 theta upon 4 plus Tan square 2 theta Right So what kind of a substitution? Well, should I go for I should substitute tan of 2 theta as you yes So secant square 2 theta into 2 d theta becomes your du Right in other words your numerator which is secant square 2 theta d theta becomes half of du. Yes Half of du. So let's put in the values over here So half of du so that will become half outside Du on the top and we have 4 plus u square and what happens to the limits of integration See when theta is 0 u is going to be a 0 But when theta is going to be a pi by 4 u is going to go all the way till infinity Correct. So you end up integrating 8 by pi from 0 to infinity Du upon 4 plus u square now This is a standard integral, right? So this integral is inspired by a standard integral which is Integral of 1 by a square plus x square which is 1 by a tan inverse x by a Right, so we'll use this expression. So let's use this in our given integral. So my role of a is been played by 2 so here my a is 2 and of course your x is you right So if I use this formula, I'm going to get 1 by 2 Tan inverse of u by 2, right and you have to put the limits of integration So when you put an infinity, this is clearly going to be Tan inverse of a very large number, which is going to be pi by 2 Okay, and when you put a 0 you end up getting a 0, right? So basically it simplifies to 8 by pi times pi by 4, which is clearly 2 So the answer to this question the area under this curve is actually 2 square units Okay, so this answer this answer becomes 2 square units Thank you so much for watching. Bye. Bye. Stay safe and stay healthy