Answer:
To find the exact area of the surface obtained by rotating the curve y = 5 - x about the x-axis over the interval 3 ≤ x ≤ 5, you can use the formula for the surface area of revolution. The formula for the surface area of revolution of a curve y = f(x) over the interval [a, b] rotated about the x-axis is given by:
\[A = 2π\int_{a}^{b} f(x)\sqrt{1 + (f'(x))^2} dx\]
In this case, \(f(x) = 5 - x\), and we need to find \(f'(x)\), which is the derivative of \(f(x)\):
\[f'(x) = -1\]
Now, we can plug these values into the formula and calculate the surface area:
\[A = 2π\int_{3}^{5} (5 - x)\sqrt{1 + (-1)^2} dx\]
Simplify:
\[A = 2π\int_{3}^{5} (5 - x)\sqrt{2} dx\]
Now, integrate with respect to x:
\[A = 2π\sqrt{2} \int_{3}^{5} (5 - x) dx\]
\[A = 2π\sqrt{2} \left[5x - \frac{x^2}{2}\right]_{3}^{5}\]
Now, evaluate the definite integral at the upper and lower limits:
\[A = 2π\sqrt{2} \left[(5(5) - \frac{5^2}{2}) - (5(3) - \frac{3^2}{2})\right]\]
\[A = 2π\sqrt{2} \left[(25 - \frac{25}{2}) - (15 - \frac{9}{2})\right]\]
\[A = 2π\sqrt{2} \left[\frac{25}{2} - \frac{15}{2} - \frac{25}{2} + \frac{9}{2}\right]\]
Now, simplify:
\[A = 2π\sqrt{2} \left[\frac{25 - 15 - 25 + 9}{2}\right]\]
\[A = 2π\sqrt{2} \left[\frac{-6}{2}\right]\]
\[A = -6π\sqrt{2}\]
So, the exact surface area of the surface obtained by rotating the curve y = 5 - x about the x-axis over the interval 3 ≤ x ≤ 5 is \(-6π\sqrt{2}\) square units.
Explanation: