Question

\(\int _0^3\sqrt{9-x^2}dx =\)

• A. \(\frac{3π}{4}\)
• B. \(\frac{9π}{4}\)
• C. \(\frac{7π}{4}\)
• D. \(\frac{5π}{4}\)

Hint:

The curve is the upper half of x^2 + y^2 = 9; the limits cover a quarter of the circle.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The expression \(y = \sqrt{9 - x^2}\) can be squared to give \(x^2 + y^2 = 9\). This is a circle with centre at the origin and radius \(3\). The positive root means only the upper half, where \(y \ge 0\).

Step 2: Key Formula or Approach:
A definite integral of a positive function equals the area under its curve. The area of a full circle is \(\pi r^2\).

Step 3: Detailed Explanation:
The limits \(x = 0\) to \(x = 3\) select the right half of the upper semicircle. That is the quarter of the circle lying in the first quadrant, between the axes and the arc.
Area of the full circle: \(\pi(3)^2 = 9\pi\).
Area of one quarter:
\[ \frac{1}{4}\times 9\pi = \frac{9\pi}{4} \]
The other options are wrong because only \(\frac{9\pi}{4}\) is exactly one quarter of \(9\pi\). Option (A), \(\frac{3\pi}{4}\), is only \(\frac{1}{12}\) of the circle. Option (C), \(\frac{7\pi}{4}\), is \(\frac{7}{36}\) of it, and option (D), \(\frac{5\pi}{4}\), is \(\frac{5}{36}\) of it. None of these fractions matches a quarter-circle region.

Final Answer:
The integral equals the area of a quarter circle of radius 3, which is option (B). \[ \boxed{\frac{9\pi}{4}} \]