Evaluate a double integral in polar coordinates?

The answer given is in the right direction. I will try to add to it.

So, x^2 - y^2 = (r cos^2(θ) - r sin^2(θ) ) r ; then you integrate and plug in the limits.

So the above becomes r^2 cos^2(θ) - r^2 sin^2(θ) and then you need to set up the integral. First we integrate with respect to r, here from. So,

∫(1, 2) ( r^2 cos^2(θ) - r^2 sin^2(θ)) dθ and that becomes

(cos (θ) - sin (θ) * r^3) / 3 ; plug in in for r and then you get

(cos (θ) - sin (θ) ) / 3 ; Then you are ready to integrate again with respect to θ and then ∫(cos (θ) - sin (θ) ) / 3 becomes:

(cos (θ) + sin (θ) ) / 3 and then you plug in 0° and 90°and that gives you area of 0.

Email me if you don´t understand.

do you like the section ... then what's the function? x^2 - y^2 is in simple terms an expression .... think we would like ? ? (x^2 - y^2) dA all of us understand that x^2 - y^2 = r^2 cos^2(?) - r^2 sin^2(?) for this reason we've ?(0, ?/2) ?(a million, 2) r^2 cos(2?) r dr d? = ?(0, ?/2) ?(a million, 2) r^3 cos(2?) dr d? it fairly is now greater handy to evaluate ....

do you need the area ... then what is the function?

x^2 - y^2 is simply an expression ....

suppose we want

∫ ∫ (x^2 - y^2) dA

we know that

x^2 - y^2 = r^2 cos^2(θ) - r^2 sin^2(θ)

thus we have

∫(0, π/2) ∫(1, 2) r^2 cos(2θ) r dr dθ

= ∫(0, π/2) ∫(1, 2) r^3 cos(2θ) dr dθ

this is now easier to evaluate ....