Understanding the Area Bounded by the Curve \(y=\sqrt{16-x^2}\) and the X-axis

The question asks for the area bounded by the curve defined by the equation \(y=\sqrt{16-x^2}\), the condition \(y \ge 0\), and the x-axis.

First, let's analyze the given equation \(y=\sqrt{16-x^2}\). Since the square root must be non-negative, the condition \(y \ge 0\) is naturally included by the definition of the square root function here, provided \(16-x^2 \ge 0\).

Squaring both sides of the equation \(y=\sqrt{16-x^2}\) (keeping in mind \(y \ge 0\)) gives us:

\(y^2 = 16 - x^2\)

Rearranging the terms, we get:

\(x^2 + y^2 = 16\)

This is the standard equation of a circle centered at the origin \((0, 0)\) with a radius squared equal to 16. Therefore, the radius of the circle is \(r = \sqrt{16} = 4\).

Since the original equation is \(y=\sqrt{16-x^2}\) (which implies \(y \ge 0\)), the graph of this equation represents only the upper half of the circle \(x^2 + y^2 = 16\). The upper half of the circle extends from \(x = -4\) to \(x = 4\).

The boundaries mentioned in the problem are:

• The curve \(y=\sqrt{16-x^2}\) (the upper semi-circle).
• The condition \(y \ge 0\) (already covered by the curve's definition).
• The x-axis (which is the line \(y=0\)).

The region bounded by the upper semi-circle \(y=\sqrt{16-x^2}\) and the x-axis (\(y=0\)) is precisely the area of the upper semi-circle of radius 4.

Calculating the Area of the Semi-Circle

The area of a full circle with radius \(r\) is given by the formula \(\text{Area}_{\text{circle}} = \pi r^2\).

The area of a semi-circle is half the area of the corresponding full circle. So, the area of the semi-circle with radius \(r\) is \(\text{Area}_{\text{semi-circle}} = \frac{1}{2} \pi r^2\).

In this problem, the radius of the semi-circle is \(r = 4\).

Let's calculate the area:

\(\text{Area} = \frac{1}{2} \pi (4)^2\)

\(\text{Area} = \frac{1}{2} \pi (16)\)

\(\text{Area} = 8\pi\)

The area bounded by the curve \(y=\sqrt{16-x^2}\), \(y \ge 0\), and the x-axis is \(8\pi\) square units.

Alternatively, this area could be found using integration:

\(\text{Area} = \int_{-4}^{4} \sqrt{16-x^2} \, dx\)

This integral represents the area under the curve \(y=\sqrt{16-x^2}\) from \(x = -4\) to \(x = 4\), which corresponds to the area of the upper semi-circle of radius 4. Evaluating this integral using a trigonometric substitution \(x = 4\sin\theta\) would also yield the result \(8\pi\).

Conclusion

The area bounded by \(y=\sqrt{16-x^2}\), \(y \ge 0\) and the x-axis is the area of a semi-circle with radius 4.

The calculated area is \(8\pi\) square units.

Revision Table: Area Under Curves and Geometric Shapes

Additional Information: Area Calculation Methods

Calculating the area bounded by curves is a fundamental concept in calculus, specifically in integral calculus. There are two main approaches:

• Geometric Formulas: If the bounded region forms a standard geometric shape (like a circle, semi-circle, rectangle, triangle), you can directly use the known area formula for that shape. This is often simpler when applicable, as seen in this problem where the region is a semi-circle.
• Integration: For more complex shapes that don't correspond to standard geometric figures, integration is used. The area bounded by a curve \(y = f(x)\), the x-axis, and the vertical lines \(x=a\) and \(x=b\) (where \(f(x) \ge 0\) on \([a,b]\)) is given by the definite integral \(\int_a^b f(x) \, dx\). If the curve is defined in terms of \(x\) as a function of \(y\), i.e., \(x=g(y)\), the area bounded by the curve, the y-axis, and horizontal lines \(y=c\) and \(y=d\) is given by \(\int_c^d g(y) \, dy\).

In this specific problem, recognizing that the equation represents a semi-circle allows for a straightforward solution using the geometric area formula, avoiding the need for integral calculus.