Area between A∩B

[rettore84]

Hello to all,

It looks a simples question but I cannot get the correct answer:

The area between A∩B, where:


A={ (x,y) ∈ R^2: 0<=x<=pi/2, 0<=y<=cos x}

B={ (x,y) ∈ R^2: 0<=x<=pi/2, sin x<=y<=1}

My answer is 2 - (pi/2). Correct answer is sqrt(2) -1

Thank you.

 

[Deleted member 4993]

Hello to all,

It looks a simples question but I cannot get the correct answer:

The area between A∩B, where:


A={ (x,y) ∈ R^2: 0<=x<=pi/2, 0<=y<=cos x}

B={ (x,y) ∈ R^2: 0<=x<=pi/2, sin x<=y<=1}

My answer is 2 - (pi/2). Correct answer is sqrt(2) -1

Thank you.

How did you get your answer?

Please share your work.

 

[rettore84]

I calculate A and B each one as a double integral and then subtract A - B.

A--> first integrate 1dy from 0 to cos(x). Results cos(x), then I integrate cos(x)dx from 0 to pi/2. Results 1.

B--> first integrate 1dy from sin(x) to 1. Results 1 - sen(x), then I integrate (1-sin(x))dx from 0 to pi/2. Results pi/2 -1.

A - B = 2 - (pi/2).

Am I doing correctly?

 

[Deleted member 4993]

Watch those limits carefully!

Area = \(\displaystyle \displaystyle{\int_0^{\dfrac{\pi}{4}}[cos(x)-sin(x)] dx}\)

Area = \(\displaystyle \left [\displaystyle{sin(\dfrac{\pi}{4}) + cos(\dfrac{\pi}{4}) - 1}\right ]\)

 

[rettore84]

Thank you Subhotosh Khan, now I understand. My aproach was completely wrong. It is a simple area integration with the limits being from 0 to the intersection of cosx=sinx. Am I correct now?