Let's solve this step by step: --- ### **Given:** - The curve: \( xy = 3 \) - Interval: \( 4 \leq y \leq 7 \) - Revolving **about the y-axis** --- ### **Step 1: Express \( x \) as a function of \( y \)** Since \( xy = 3 \): \[ x = \frac{3}{y} \] --- ### **Step 2: Formula for Surface Area of Revolution (about the y-axis)** The surface area \( S \) when revolving \( x = f(y) \) from \( y = a \) to \( y = b \) about the y-axis is: \[ S = 2\pi \int_{a}^{b} x \sqrt{1 + \left( \frac{dx}{dy} \right)^2} \, dy \] --- ### **Step 3: Find \( \frac{dx}{dy} \)** \[ x = \frac{3}{y} \] \[ \frac{dx}{dy} = -\frac{3}{y^2} \] --- ### **Step 4: Plug everything into the surface area formula** \[ S = 2\pi \int_{4}^{7} \frac{3}{y} \sqrt{1 + \left(-\frac{3}{y^2}\right)^2} \, dy \] \[ = 2\pi \int_{4}^{7} \frac{3}{y} \sqrt{1 + \frac{9}{y^4}} \, dy \] --- ### **Step 5: Final Answer** \[ \boxed{ S = 2\pi \int_{4}^{7} \frac{3}{y} \sqrt{1 + \frac{9}{y^4}} \, dy } \] This is the required integral for the area of the surface. (You can use technology to evaluate the definite integral numerically if needed.)

Let's compute the surface area step-by-step numerically. --- ### **Given:** - \( xy = 3 \) - Interval for \( y \): \( 4 \le y \le 7 \) - \( x = \frac{3}{y} \) - \( \frac{dx}{dy} = -\frac{3}{y^2} \) --- ### **Surface Area Formula:** \[ S = 2\pi \int_{4}^{7} x \sqrt{1 + \left( \frac{dx}{dy} \right)^2} \, dy \] Substitute \( x = \frac{3}{y} \) and \( \frac{dx}{dy} = -\frac{3}{y^2} \): \[ S = 2\pi \int_{4}^{7} \frac{3}{y} \sqrt{1 + \frac{9}{y^4}} \, dy \] --- ### **Step 1: Simplify the integrand** \[ \sqrt{1 + \frac{9}{y^4}} = \sqrt{\frac{y^4 + 9}{y^4}} = \frac{\sqrt{y^4 + 9}}{y^2} \] So, \[ S = 2\pi \int_{4}^{7} \frac{3}{y} \times \frac{\sqrt{y^4 + 9}}{y^2} \, dy = 2\pi \int_{4}^{7} \frac{3 \sqrt{y^4 + 9}}{y^3} \, dy \] --- ### **Step 2: Write the integral explicitly** \[ S = 6\pi \int_{4}^{7} \frac{\sqrt{y^4 + 9}}{y^3} \, dy \] --- ### **Step 3: Numerical evaluation** We evaluate: \[ I = \int_{4}^{7} \frac{\sqrt{y^4 + 9}}{y^3} \, dy \] Using numerical methods (such as Simpson's rule or a calculator), let's approximate this integral. --- ### **Approximate the integral \( I \):** Evaluate the integrand at key points: | \( y \) | \( \text{Integrand} = \frac{\sqrt{y^4 + 9}}{y^3} \) | |---------|--------------------------------------------------| | 4 | \( \frac{\sqrt{4^4 + 9}}{4^3} = \frac{\sqrt{256 + 9}}{64} = \frac{\sqrt{265}}{64} \approx \frac{16.28}{64} \approx 0.254 \) | | 5 | \( \frac{\sqrt{625 + 9}}{125} = \frac{\sqrt{634}}{125} \approx \frac{25.19}{125} \approx 0.201 \) | | 6 | \( \frac{\sqrt{1296 + 9}}{216} = \frac{\sqrt{1305}}{216} \approx \frac{36.14}{216} \approx 0.167 \) | | 7 | \( \frac{\sqrt{2401 + 9}}{343} = \frac{\sqrt{2410}}{343} \approx \frac{49.09}{343} \approx 0.143 \) | Approximate integral using Simpson's rule with these four points: \[ I \approx \frac{h}{3} \left[ f(y_0) + 4f(y_1) + 2f(y_2) + 4f(y_3) + f(y_4) \right] \] where: - \( y_0=4 \), \( y_1=5 \), \( y_2=6 \), \( y_3=7 \) and \[ h = y_1 - y_0 = 1 \] The Simpson's rule (with 3 intervals): \[ I \approx \frac{1}{3} \left[ 0.254 + 4 \times 0.201 + 2 \times 0.167 + 4 \times 0.143 + \text{(extra point at } y=7) \right] \] But since we only have 4 points, a simple trapezoidal approximation suffices for an estimate: \[ I \approx \frac{h}{2} [f(4) + 2f(5) + 2f(6) + f(7)] \] Alternatively, for a rough estimate, we can average the integrand: \[ I \approx \text{average of integrand} \times (7 - 4) = \left( \frac{0.254 + 0.201 + 0.167 + 0.143}{4} \right) \times 3 \] Average: \[ \frac{0.254 + 0.201 + 0.167 + 0.143}{4} = \frac{0.765}{4} \approx 0.191 \] Multiply by interval length: \[ I \approx 0.191 \times 3 = 0.573 \] --- ### **Step 4: Final surface area** Recall: \[ S = 6\pi \times I \approx 6 \times 3.1416 \times 0.573 \approx 18.8496 \times 0.573 \approx 10.8 \] --- ### **### **Final answer:** \[ \boxed{ \text{Surface Area} \approx \boxed{10.8} } \] **Note:** More precise numerical integration (using a calculator or software) would give a slightly more accurate result, but this estimate is close enough for practical purposes.