Multivariable Calculus Question of the Day

1. $\int_{-\pi/2}^{\pi/2} \int_{1}^{1+\cos\theta} 1 \,dr\,d\theta$
2. $\int_{-\pi/2}^{\pi/2} \int_{1}^{1+\cos\theta} r \,dr\,d\theta$ (correct answer)
3. $\int_{0}^{2\pi} \int_{1}^{1+\cos\theta} r \,dr\,d\theta$
4. $\int_{-\pi/2}^{\pi/2} \int_{0}^{1+\cos\theta} r \,dr\,d\theta$

Explanation: When setting up double integrals in polar coordinates to find area, you need to remember two key components: the correct bounds of integration and the proper area element, which is $r \, dr \, d\theta$ (not just $dr \, d\theta$). For this problem, you're finding the area between two curves in the right half-plane. The region $R$ is bounded by the inner circle $r = 1$ and outer cardioid $r = 1 + \cos\theta$. Since you want only the right half-plane ($x \geq 0$), you need $\theta$ to range from $-\pi/2$ to $\pi/2$. For each fixed angle $\theta$ in this range, $r$ varies from the inner boundary $r = 1$ to the outer boundary $r = 1 + \cos\theta$. Answer B correctly captures both requirements: $\int_{-\pi/2}^{\pi/2} \int_{1}^{1+\cos\theta} r \,dr\,d\theta$ Answer A uses the wrong area element ($1$ instead of $r$) – this would give you area in rectangular coordinates, not polar. Answer C has the correct area element but wrong $\theta$ bounds ($0$ to $2\pi$ instead of $-\pi/2$ to $\pi/2$), which would give you the entire region around both curves, not just the right half-plane. Answer D has the wrong inner $r$-bound (starting from $0$ instead of $1$), which would include the area inside the circle $r = 1$ rather than just the region between the two curves. Remember: in polar coordinates, area integrals always need the factor $r$ in the integrand, and carefully determine your bounds by visualizing the region you want.