I am currently taking my first calculus class as a first semester maths student. The following questions is from the book Calculus: Early Transcendentals, on the chapter regarding applications of double integrals. Question is written as stated: Find the polar moment of inertia $I_0$ of the indicated lamina. The region bounded by the right-hand loop of the lemniscate $r^2=cos(2\theta);\delta(x,y)=r^2$.
The given laminate can be seen here.
From what i understand the inertia can be described as the following in polar coordinates: $$I_0=\iint_R r^2\delta(x,y)r \,dr\,d\theta$$ If i were to plug in the given information, i would find the following iterated integral: $$\int_{-\pi/2}^{\pi/2}\int_{0}^{\sqrt{cos(2\theta)}} r^2r^2rdrd\theta$$ Here i define the bounds for r as the positive solution to the given $r^2$ equation. Meanwhile the $\theta$ limits are given by the question, as the regions is in the first and fourth quadrant.
However, I realise that this is wrong as it computes to 0. Any tips on how to move further along with this? Any help is appreciated, thank you in advance.
The limits for $\theta$ should be $$ -\frac{\pi}{4} \leq \theta \leq \frac{\pi}{4}. $$ Note that the right part of the lemniscate lies between the lines $y=x$ and $y=-x$. I believe the picture you included isn't correct.
Also, from an algebraic point of view we see the limits weren't correct. Note that $\cos(2\theta)$ becomes negative when $\pi/4<\theta \leq \pi/2$ or $ -\pi/2\leq \theta < -\pi/4$, so $\sqrt{\cos(2\theta)}$ is not defined for these values.