Question

Evaluate: \(\int_0^3 \sqrt{9 - x^2} \, dx\).

Hint:

Geometry shortcut: \(\int_0^r \sqrt{r^2 - x^2} \, dx\) is simply the area of a quadrant of a circle, which is \(\frac{1}{4} \pi r^2\). Here \(r=3\), so Area \(= \frac{1}{4} \pi (3)^2 = \frac{9\pi}{4}\). This is much faster than integration!

Answer 1

Step 1: Understanding the Question:
The integral represents the area of a portion of a circle \( x^2 + y^2 = 9 \) in the first quadrant.
Step 2: Key Formula or Approach:
Standard integral: \( \int \sqrt{a^2-x^2} dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a} \).
Step 3: Detailed Explanation:
\[ I = \int_0^3 \sqrt{3^2 - x^2} \, dx \]
Applying the formula:
\[ I = \left[ \frac{x}{2}\sqrt{9-x^2} + \frac{9}{2}\sin^{-1}\frac{x}{3} \right]_0^3 \]
Upper limit (\( x=3 \)): \( 0 + \frac{9}{2}\sin^{-1}(1) = \frac{9}{2} \cdot \frac{\pi}{2} = \frac{9\pi}{4} \).
Lower limit (\( x=0 \)): \( 0 + \frac{9}{2}\sin^{-1}(0) = 0 \).
Result \( = \frac{9\pi}{4} - 0 = \frac{9\pi}{4} \).
Step 4: Final Answer:
The integral value is \( 9\pi/4 \).