Step 1: Identifying the Geometric Properties

The given equation is $\frac{x^2}{16} + \frac{y^2}{9} = 1$. This is the standard form of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where $a^2 = 16$ and $b^2 = 9$.

• Semi-major axis ($a$): $\sqrt{16} = 4$ units.
• Semi-minor axis ($b$): $\sqrt{9} = 3$ units.
• Center: $(0,0)$.

Step 2: Graphical Representation

Step 3: Setting up the Integral

Due to the symmetry of the ellipse about both the $x$-axis and $y$-axis, the total area $A$ is four times the area of the region in the first quadrant. [By Symmetry Principle]

From $\frac{x^2}{16} + \frac{y^2}{9} = 1$, we isolate $y$ for the first quadrant ($y \ge 0$):

$\frac{y^2}{9} = 1 - \frac{x^2}{16} \implies y^2 = \frac{9}{16}(16 - x^2) \implies y = \frac{3}{4}\sqrt{16 - x^2}$

The area in the first quadrant is $\int_{0}^{4} y \, dx$. Thus, the total area $A$ is:

$A = 4 \int_{0}^{4} \frac{3}{4} \sqrt{4^2 - x^2} \, dx = 3 \int_{0}^{4} \sqrt{4^2 - x^2} \, dx$

Step 4: Evaluating the Definite Integral

Using the standard integral formula: $\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1}(\frac{x}{a})$

$A = 3 \left[ \frac{x}{2}\sqrt{16 - x^2} + \frac{16}{2} \sin^{-1}\left(\frac{x}{4}\right) \right]_{0}^{4}$

$A = 3 \left[ \left( \frac{4}{2}\sqrt{16 - 16} + 8 \sin^{-1}(1) \right) - \left( 0 + 8 \sin^{-1}(0) \right) \right]$

$A = 3 \left[ 0 + 8\left(\frac{\pi}{2}\right) - 0 \right]$

$A = 3 \times 4\pi = 12\pi$

Summary of Calculation

Final Solution: The area of the region bounded by the ellipse is $12\pi$ square units.

Step 1: Analyzing the Equation of the Ellipse

The given equation of the ellipse is $\frac{x^2}{4} + \frac{y^2}{9} = 1$. This is in the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where:

• $a^2 = 4 \implies a = 2$ (Semi-major axis along the $x$-axis is $2$)
• $b^2 = 9 \implies b = 3$ (Semi-major axis along the $y$-axis is $3$)

[Per standard conic section definitions, since $b > a$, the ellipse is elongated along the $y$-axis.]

Step 2: Expressing $y$ in terms of $x$

To find the area using integration, we isolate $y$ from the equation:

$\frac{y^2}{9} = 1 - \frac{x^2}{4}$

$y^2 = 9 \left(1 - \frac{x^2}{4}\right) = \frac{9}{4}(4 - x^2)$

$y = \pm \frac{3}{2}\sqrt{4 - x^2}$

Since the ellipse is symmetric about both axes, the total area $A$ is 4 times the area of the region in the first quadrant where $x$ ranges from $0$ to $2$.

Step 3: Setting up the Definite Integral

$A = 4 \int_{0}^{2} y \, dx = 4 \int_{0}^{2} \frac{3}{2} \sqrt{2^2 - x^2} \, dx$

$A = 6 \int_{0}^{2} \sqrt{2^2 - x^2} \, dx$

Step 4: Applying the Integration Formula

Using the standard integration formula $\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)$:

$A = 6 \left[ \frac{x}{2}\sqrt{4 - x^2} + \frac{4}{2}\sin^{-1}\left(\frac{x}{2}\right) \right]_{0}^{2}$

$A = 6 \left[ \left( \frac{2}{2}\sqrt{4 - 4} + 2\sin^{-1}(1) \right) - \left( 0 + 2\sin^{-1}(0) \right) \right]$

$A = 6 \left[ 0 + 2(\frac{\pi}{2}) - 0 \right]$

$A = 6 \pi$

Visual Representation of the Ellipse

Summary of Calculation

Final Solution: The area of the region bounded by the ellipse is $6\pi$ square units.

Step 1: Analyzing the Geometric Constraints

The given equation is $x^2 + y^2 = 4$, which represents a circle centered at the origin $(0, 0)$ with radius $r = \sqrt{4} = 2$ units [By the standard form of circle equation $(x-h)^2 + (y-k)^2 = r^2$].

The boundary conditions are:

• The circle: $y^2 = 4 - x^2 \implies y = \sqrt{4 - x^2}$ (taking the positive root as we are constrained to the first quadrant).
• The vertical lines: $x = 0$ (the $y$-axis) and $x = 2$ (the rightmost edge of the circle).

Step 2: Defining the Integral Setup

The area $A$ in the first quadrant bounded by the curve $f(x) = \sqrt{4 - x^2}$ from $x = 0$ to $x = 2$ is given by the definite integral:

$A = \int_{0}^{2} \sqrt{4 - x^2} \, dx$

Step 3: Evaluating the Integral

Using the standard integration formula $\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1}\left(\frac{x}{a}\right) + C$:

Substitute $a = 2$:

$A = \left[ \frac{x}{2}\sqrt{4 - x^2} + \frac{4}{2} \sin^{-1}\left(\frac{x}{2}\right) \right]_{0}^{2}$

$A = \left[ \frac{x}{2}\sqrt{4 - x^2} + 2 \sin^{-1}\left(\frac{x}{2}\right) \right]_{0}^{2}$

Applying the Fundamental Theorem of Calculus:

$A = \left( \frac{2}{2}\sqrt{4 - 2^2} + 2 \sin^{-1}\left(\frac{2}{2}\right) \right) - \left( \frac{0}{2}\sqrt{4 - 0^2} + 2 \sin^{-1}\left(\frac{0}{2}\right) \right)$

$A = (1 \cdot 0 + 2 \cdot \sin^{-1}(1)) - (0 + 2 \cdot 0)$

$A = 2 \cdot \left(\frac{\pi}{2}\right) = \pi$

Step 4: Geometric Visualization

Final Solution: The area is $\pi$ square units.

Given Variables & Initial Setup

The region is bounded by:

• The parabola: $y^2 = 4x$
• The $y$-axis: $x = 0$
• The horizontal line: $y = 3$

To find the area with respect to the $y$-axis, we express $x$ in terms of $y$ from the given equation of the parabola:

$x = \frac{y^2}{4}$ [By algebraic rearrangement]

Step 1: Establishing the Limits of Integration

The region starts from the origin where $y = 0$ and extends up to the line $y = 3$. Therefore, the definite integral limits are $[0, 3]$.

Step 2: Defining the Integral for Area

The area $A$ under the curve bounded by the $y$-axis is given by the integral of $x$ with respect to $y$:

$A = \int_{c}^{d} x \, dy = \int_{0}^{3} \frac{y^2}{4} \, dy$

Step 3: Calculating the Integral

Applying the Power Rule for integration, $\int y^n \, dy = \frac{y^{n+1}}{n+1}$:

$A = \frac{1}{4} \int_{0}^{3} y^2 \, dy$

$A = \frac{1}{4} \left[ \frac{y^3}{3} \right]_{0}^{3}$ [By the Fundamental Theorem of Calculus]

$A = \frac{1}{4} \left( \frac{3^3}{3} - \frac{0^3}{3} \right)$

$A = \frac{1}{4} \left( \frac{27}{3} \right)$

$A = \frac{1}{4} (9)$

$A = \frac{9}{4} \text{ sq. units}$

Visual Representation

Final Solution: The area of the region is $\frac{9}{4}$ or $2.25 \text{ square units}$.