Using the definite integration area of the circle with radius 4 unit is______

Options

A. `5pi ` sq . Untis
B. 4sq. Units
C. 16 sq. units
D. `16pi sq. `units

Answer

To find the area of a circle with a radius of 4 units using definite integration, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Circle's Equation**:
The equation of a circle with center at the origin (0,0) and radius \( r \) is given by:
\[ x^2 + y^2 = r^2 \]
For our case, \( r = 4 \), so the equation becomes:
\[ x^2 + y^2 = 16 \]

2. **Express \( y \) in Terms of \( x \)**:
We can express \( y \) as:
\[ y = \sqrt{16 - x^2} \]
This represents the upper half of the circle.

3. **Set Up the Integral**:
To find the area of the upper half of the circle, we will integrate from \( x = 0 \) to \( x = 4 \):
\[ \text{Area}_{\text{upper half}} = \int_{0}^{4} \sqrt{16 - x^2} \, dx \]

4. **Use a Trigonometric Substitution**:
Let \( x = 4 \sin(t) \). Then, \( dx = 4 \cos(t) \, dt \). The limits change as follows:
- When \( x = 0 \), \( t = 0 \)
- When \( x = 4 \), \( t = \frac{\pi}{2} \)

Now, substitute into the integral:
\[ \int_{0}^{\frac{\pi}{2}} \sqrt{16 - (4 \sin(t))^2} \cdot 4 \cos(t) \, dt \]
Simplifying the integrand:
\[ = \int_{0}^{\frac{\pi}{2}} \sqrt{16(1 - \sin^2(t))} \cdot 4 \cos(t) \, dt \]
\[ = \int_{0}^{\frac{\pi}{2}} \sqrt{16 \cos^2(t)} \cdot 4 \cos(t) \, dt \]
\[ = \int_{0}^{\frac{\pi}{2}} 4 \cdot 4 \cos^2(t) \, dt \]
\[ = 16 \int_{0}^{\frac{\pi}{2}} \cos^2(t) \, dt \]

5. **Integrate \( \cos^2(t) \)**:
Using the identity \( \cos^2(t) = \frac{1 + \cos(2t)}{2} \):
\[ = 16 \int_{0}^{\frac{\pi}{2}} \frac{1 + \cos(2t)}{2} \, dt \]
\[ = 8 \left[ t + \frac{\sin(2t)}{2} \right]_{0}^{\frac{\pi}{2}} \]
Evaluating the limits:
\[ = 8 \left( \frac{\pi}{2} + 0 - (0 + 0) \right) = 8 \cdot \frac{\pi}{2} = 4\pi \]

6. **Calculate the Total Area**:
Since this is the area of the upper half of the circle, the total area of the circle is:
\[ \text{Total Area} = 2 \cdot 4\pi = 8\pi \]

### Final Answer:
The area of the circle with radius 4 units is:
\[ \text{Area} = 16\pi \]