simplycoder

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7 years ago

Show us your work buddy..

In polar form put r=under root x2+y2 and tan@ is y/x

so the required eqn are

x2+ y2=a2

(x-a)2+y2=a2

so now we can draw two circles

the required area will be 2(area of upper curve -lower curve)

the upper curve is second circle nd lower curve is first one.

I took Limit of integrtn as 0 to a/2

but I dont get the answer.

so the required eqn are

x2+ y2=a2

(x-a)2+y2=a2

so now we can draw two circles

the required area will be 2(area of upper curve -lower curve)

the upper curve is second circle nd lower curve is first one.

I took Limit of integrtn as 0 to a/2

but I dont get the answer.

simplycoder

Branch Unspecified

7 years ago

I will give you a hint here, draw the figure on paper and see whether you have correct limits.

simplycoder

Branch Unspecified

7 years ago

Can you put a pic that might show how did you get those limits?

How did you come with limits from a/2 to a and other stuff?

What is the center of the two circles. If you can get this right then only we are on correct path.

Did you try to find the second integral limits?

What co-ordinate system did you choose? what is the second integral?

Think about it conceptually and not something to solve and match the answer.

BTW if you have the final answer, then post that too, so that when I finally explain there wouldnt be any issues as there are multiple equivalent forms of same answer.

How did you come with limits from a/2 to a and other stuff?

What is the center of the two circles. If you can get this right then only we are on correct path.

Did you try to find the second integral limits?

What co-ordinate system did you choose? what is the second integral?

Think about it conceptually and not something to solve and match the answer.

BTW if you have the final answer, then post that too, so that when I finally explain there wouldnt be any issues as there are multiple equivalent forms of same answer.

as the equation is in polar form so first convert it to cartesian co ordinates . As i have explained previously. So the centres comes out to be 0,0 and a,0. And radius for both are a . Now when we draw both circles intersect on x axis . So that point is a/2 ,0.

Now you tell me what to do now.

This is gates question. Answer is 1.77 a2.

Now you tell me what to do now.

This is gates question. Answer is 1.77 a2.

I_am_Kunal :)

Branch Unspecified

7 years ago

The circles do not intersect on the x-axis. You are right about the centers. But you should recheck the points of intersections. The circles intersect at two points. X co-ordinate of both of which is a/2. Y co-ordinate is not 0.RVigneshas the equation is in polar form so first convert it to cartesian co ordinates . As i have explained previously. So the centres comes out to be 0,0 and a,0. And radius for both are a . Now when we draw both circles intersect on x axis . So that point is a/2 ,0.

Now you tell me what to do now.

This is gates question. Answer is 1.77 a2.

simplycoder

Branch Unspecified

7 years ago

Logically think about it.

Since we are interested in polar co-ordinates, we only want to know more in terms of @ and a.

So you got a (Though I think its incorrect) Now search for limits in terms of @.

Then double integrate, still if you dont get, I shall upload the answer tomorrow.

Also I donot agree with your answer and so if there is any method along it, post that aswell. I am getting 1.228(a^2). How?? Will let you know by tomorrow.

Since we are interested in polar co-ordinates, we only want to know more in terms of @ and a.

So you got a (Though I think its incorrect) Now search for limits in terms of @.

Then double integrate, still if you dont get, I shall upload the answer tomorrow.

Also I donot agree with your answer and so if there is any method along it, post that aswell. I am getting 1.228(a^2). How?? Will let you know by tomorrow.

simplycoder

Branch Unspecified

7 years ago

This is a rough sketch of what your graph should look like.

Solve for @ and you get that it intersects in two points pi/3 and -pi/3.

The red area is from (@=-pi/3 to pi/3) and (r=0 to a).

The blue area follows the r=2cos@ circle.

And there are two such of them. So no integrate

I(r=0 to a , @=-pi/3 to pi/3)rdrd@ ................................(1)

2*I(r=0 to cos2@ , @=pi/3 to pi/2)rdrd@ ................................(2)

Add both the integrals, you would get the answer.

Solve for @ and you get that it intersects in two points pi/3 and -pi/3.

The red area is from (@=-pi/3 to pi/3) and (r=0 to a).

The blue area follows the r=2cos@ circle.

And there are two such of them. So no integrate

I(r=0 to a , @=-pi/3 to pi/3)rdrd@ ................................(1)

2*I(r=0 to cos2@ , @=pi/3 to pi/2)rdrd@ ................................(2)

Add both the integrals, you would get the answer.

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