Question

Sketch the curve described by the equation $\{(x, y) : 9x^2 + 16y^2 = 144\}$ and find the exact total area of the closed region enclosed by it using definite integration.

Hint:

The total area of any standard ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ can be calculated directly using the formula $\text{Area} = \pi a b$. For this problem, $\text{Area} = \pi \times 4 \times 3 = 12\pi$. Use this formula to check your answer instantly!

Answer 1

Concept: The equation $9x^2 + 16y^2 = 144$ represents a standard horizontal ellipse centered at the origin. To analyze it, we rewrite it in standard form: \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \] The total enclosed geometric area can be calculated by integrating a thin vertical area strip $y \, dx$ across a bounding interval. Since an ellipse is perfectly symmetrical across both the $x$-axis and $y$-axis, the total area is equal to 4 times the area of the region contained within the first quadrant: \[ \text{Total Area} = 4 \int_{0}^{a} y \, dx \]

Step 1: Converting the equation to standard form to find the semi-axes.
The given equation is: \[ 9x^2 + 16y^2 = 144 \] Divide both sides of the equation by 144 to make the right side equal to 1: \[ \frac{9x^2}{144} + \frac{16y^2}{144} = 1 \implies \frac{x^2}{16} + \frac{y^2}{9} = 1 \] Rewriting this with squared denominators: \[ \frac{x^2}{4^2} + \frac{y^2}{3^2} = 1 \] Comparing this to the standard form shows that the curve is an ellipse with:
• Semi-major axis length $a = 4$ along the $x$-axis.
• Semi-minor axis length $b = 3$ along the $y$-axis. The ellipse intersects the coordinate axes at the vertices $(\pm 4, 0)$ and $(0, \pm 3)$.

Step 2: Isolating the variable $y$ as an explicit function of $x$.
Let us solve for $y$ from our standard form equation: \[ \frac{y^2}{9} = 1 - \frac{x^2}{16} \implies \frac{y^2}{9} = \frac{16 - x^2}{16} \] Multiply both sides by 9: \[ y^2 = \frac{9}{16}(16 - x^2) \] Taking the positive square root for our first-quadrant calculation analysis ($y \geq 0$): \[ y = \frac{3}{4}\sqrt{16 - x^2} \]

Step 3: Setting up the definite integral for the area.
Using the symmetry property, integrate from $x = 0$ to $x = 4$ and multiply the result by 4: \[ \text{Area} = 4 \int_{0}^{4} y \, dx = 4 \int_{0}^{4} \frac{3}{4}\sqrt{16 - x^2} \, dx \] Cancel out the constant coefficient fraction 4: \[ \text{Area} = 3 \int_{0}^{4} \sqrt{4^2 - x^2} \, dx \]

Step 4: Evaluating the integral using standard integration formulas.
Apply the standard trigonometric radical integration formula $\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)$: \[ \text{Area} = 3 \left[ \frac{x}{2}\sqrt{16 - x^2} + \frac{16}{2}\sin^{-1}\left(\frac{x}{4}\right) \right]_{0}^{4} \] \[ \text{Area} = 3 \left[ \frac{x}{2}\sqrt{16 - x^2} + 8\sin^{-1}\left(\frac{x}{4}\right) \right]_{0}^{4} \] Now substitute the upper boundary limit ($x = 4$) and lower boundary limit ($x = 0$):
• Substituting $x = 4$: $\frac{4}{2}\sqrt{16 - 16} + 8\sin^{-1}\left(\frac{4}{4}\right) = 0 + 8\sin^{-1}(1) = 8 \left(\frac{\pi}{2}\right) = 4\pi$
• Substituting $x = 0$: $\frac{0}{2}\sqrt{16 - 0} + 8\sin^{-1}(0) = 0 + 0 = 0$ Multiply the result by our constant factor 3: \[ \text{Area} = 3 \times (4\pi - 0) = 12\pi \text{ square units} \] Thus, the total area enclosed by the ellipse curve is exactly $12\pi$ square units.