Question

Anybody know how to find the integral of following figure?

Answer 1

I mean area of the triangle using polar co-ordinates

Answer 2

using a double integral?

Answer 3

rcostheta + rsintheta =1
r=1/(costheta +sintheta)

\[\int\limits_{0}^{\frac{ \pi }{ 2 }} \frac{ r^2 }{2 } d \theta = \frac{ 1 }{ 2 }\int\limits_{0}^{\frac{ \pi }{ 2 }} \frac{ 1 }{(\cos \theta +\sin \theta )^2} d \theta \]

Answer 4

@TuringTest yes @Algebraic! how did u got that? can you explain some more?

Answer 5

x=rcostheta
y=rsintheta

Answer 6

area = integral (r^2)/2 d theta

Answer 7

why is that?

Answer 8

the differential of of the area of a sector is given by \[dA=\frac{ 1 }{ 2 }r^2 d \theta \] so the area of the region is a definite integral \[A=\int\limits_{a}^{b}\frac{ 1 }{ 2 }r^2 d \theta \].

Answer 9

but we need to find area of triangle

Answer 10

@Algebraic! isn't that integral from 0 to pi/4

Answer 11

pi/2

Answer 12

are you in calc. 2?

Answer 13

Oh yea it's pi/2 thnks @Algebraic! ......but why dA is r^2/s d(theta)

Answer 14

because the area of a circular sector is r^2 /2 times theta

Answer 15

Finally I got it thnks guys

Answer 16

the small triangle is approximated by a sector of a circle.