Let's go step by step: --- ### (a) **Reverse the order of integration** Given: \[ \int_{0}^{\frac{\pi}{3}} \int_{3y}^{\pi} \sin(x^2) \, dx dy \] #### **Step 1: Identify the Region** The limits are: - \( y \) goes from \( 0 \) to \( \frac{\pi}{3} \) - For each \( y \), \( x \) goes from \( 3y \) to \( \pi \) So, the region \( R \) in the \(xy\)-plane is: \[ R = \left\{ (x, y) \mid 0 \leq y \leq \frac{\pi}{3},\ 3y \leq x \leq \pi \right\} \] #### **Step 2: Find New Bounds** - From \( 3y \leq x \leq \pi \), \( y \) can be written as \( y \geq 0 \) and \( y \leq \frac{x}{3} \). - But \( y \leq \frac{\pi}{3} \), so \( y \leq \min\left(\frac{x}{3}, \frac{\pi}{3}\right) \). - Also, \( x \geq 3y \implies x \geq 0 \) (since \( y \geq 0 \)), and \( x \leq \pi \). But to get the correct region, let's consider the intersection: - For \( y \leq \frac{\pi}{3} \), \( x \geq 3y \implies x \geq 0 \) up to \( x = \pi \). - For the lower bound of \( x \), as \( y \) goes from \( 0 \) up to \( \frac{\pi}{3} \), \( x \) goes from \( 0 \) up to \( \pi \). But since \( x \geq 3y \) and \( y \leq \frac{\pi}{3} \), then \( x \) starts from \( 0 \) (when \( y = 0 \)) up to \( x = \pi \), but for a fixed \( x \), \( y \) runs from \( y = 0 \) up to \( y = \frac{x}{3} \) (but not exceeding \( \frac{\pi}{3} \)). But since \( x \leq \pi \), and \( y \leq \frac{\pi}{3} \), we have \( x \) running from \( 0 \) to \( \pi \), but for \( x \), \( y \) runs from \( 0 \) to \( \min\left(\frac{x}{3}, \frac{\pi}{3}\right) \). But for \( x \) from \( 0 \) to \( \pi \), \( \frac{x}{3} \leq \frac{\pi}{3} \) when \( x \leq \pi \), so \( y \) runs from \( 0 \) to \( \frac{x}{3} \). So, the reversed order is: \[ \int_{x=0}^{\pi} \int_{y=0}^{x/3} \sin(x^2) \, dy dx \] --- ### (b) **Compute the Integral** \[ \int_{0}^{\pi} \int_{0}^{x/3} \sin(x^2) \, dy dx \] #### **Step 1: Integrate with respect to \( y \)** \[ \int_{0}^{\pi} \left[ \int_{0}^{x/3} \sin(x^2) \, dy \right] dx \] Since \(\sin(x^2)\) is **independent** of \( y \): \[ = \int_{0}^{\pi} \sin(x^2) \cdot \left[ y \right]_{y=0}^{y=x/3} dx = \int_{0}^{\pi} \sin(x^2) \cdot \frac{x}{3} dx \] \[ = \frac{1}{3} \int_{0}^{\pi} x \sin(x^2) dx \] #### **Step 2: Substitute \( u = x^2 \)** Let \( u = x^2 \implies du = 2x dx \implies x dx = \frac{du}{2} \) When \( x = 0, u = 0 \) When \( x = \pi, u = \pi^2 \) So, \[ = \frac{1}{3} \int_{x=0}^{x=\pi} x \sin(x^2) dx = \frac{1}{3} \int_{u=0}^{u=\pi^2} \sin(u) \cdot \frac{du}{2} = \frac{1}{6} \int_{0}^{\pi^2} \sin(u) du \] \[ \int \sin(u) du = -\cos(u) \] So, \[ = \frac{1}{6} \left[ -\cos(u) \right]_{0}^{\pi^2} = -\frac{1}{6} \left( \cos(\pi^2) - \cos(0) \right) \] \[ = -\frac{1}{6} \left( \cos(\pi^2) - 1 \right) \] \[ = \frac{1}{6} \left(1 - \cos(\pi^2)\right) \] --- ## **Final Answers** ### **(a) Reversed Order:** \[ \boxed{ \int_{0}^{\pi} \int_{0}^{x/3} \sin(x^2) \, dy dx } \] ### **(b) Value of the Integral:** \[ \boxed{ \frac{1}{6} \left(1 - \cos(\pi^2)\right) } \]

Reversing the order of integration involves expressing the region in terms of \(x\) first, then \(y\), by analyzing the bounds to simplify the evaluation of the integral.