Find the area of the region bounded by r^2=162cos2theta?

This is a lemniscate ("figure 8" curve).

By symmetry, it suffices to find the area of half of one of the two loops,

and then quadruple that result.

Solving for r yields r = 9√2 √(cos(2θ)).

One of these loops is traced out for θ in [-π/4, π/4], noting that cos(2θ) = 0 at the endpoints.

So, the area ∫∫ 1 dA equals

4 * ∫(θ = 0 to π/4) ∫(r = 0 to 9√2 √(cos(2θ)) 1 * (r dr dθ), converting to polar coordinates

= 4 * ∫(θ = 0 to π/4) (1/2)r^2 {for r = 0 to 9√2 √(cos(2θ)} dθ

= 4 * ∫(θ = 0 to π/4) (1/2) * 162 cos(2θ) dθ

= 162 sin(2θ) {for θ = 0 to π/4}

= 162.

I hope this helps!