# Find the area of the region bounded by the ellipse x²/16 + y²/9 = 1 **URL:** \ **Category:** Maths Question\ **Created:** [Ekim 27, 2026, 1:34öö UTC](https://sorumatik.co/t/find-the-area-of-the-region-bounded-by-the-ellipse-x-16-y-9-1/698416 "2026-10-27T01:34:13Z")\ **Posts on this page:** 2\ **Page:** 1
**Author:** ![Dksh](https://sorumatik.co/letter_avatar_proxy/v4/letter/d/8baadc/32.png) [@Dksh](https://sorumatik.co/u/Dksh)\ **Post date:** [Ekim 27, 2026, 1:34öö UTC](https://sorumatik.co/t/find-the-area-of-the-region-bounded-by-the-ellipse-x-16-y-9-1/698416/1 "2026-10-27T01:34:13Z")
Find the area of the region bounded by the ellipse \frac{x^2}{16} + \frac{y^2}{9} = 1. Choose the correct answer in the following Exercises 3 and 4. Area lying in the first quadrant and bounded by the circle x^2 + y^2 = 4 and the lines x = 0 and x = 2 is (A) \pi (B) \frac{\pi}{2} (C) \frac{\pi}{3} (D) \frac{\pi}{4} Area of the region bounded by the curve y^2 = 4x , y-axis and the line y = 3 is (A) 2 (B) \frac{9}{4} (C) \frac{9}{3} (D) \frac{9}{2} ![Find the area of the region bounded by the ellipse x²/16 + y²/9 = 1 .](https://sorumatik.co/uploads/default/original/3X/e/c/ecc779975a0e0d59780f9e1eb7388e27eb3a8a70.jpeg) ---
**Author:** ![system](https://sorumatik.co/uploads/default/original/2X/b/bfa9260de78d71e38ac0b7ec2e03a8e53e4ddcfb.png) [@system](https://sorumatik.co/u/system)\ **Post date:** [Ekim 27, 2026, 1:34öö UTC](https://sorumatik.co/t/find-the-area-of-the-region-bounded-by-the-ellipse-x-16-y-9-1/698416/2 "2026-10-27T01:34:26Z")
## Solutions to Calculus Exercises: Areas Bounded by Curves * * * ### Exercise 1: Area of the Region Bounded by the Ellipse 💡 **FORMULAS & RULES USED:** \text{Standard Equation of an Ellipse: } \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \text{Area using Integration: } A = 4 \int\_{0}^{a} y \, dx = 4 \int\_{0}^{a} \frac{b}{a}\sqrt{a^2 - x^2} \, dx \text{Standard Integral: } \int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\arcsin\left(\frac{x}{a}\right) \text{Geometric Area Formula: } A = \pi a b * * * 🧠 **STEP-BY-STEP SOLUTION:** **Step 1 — Identify the Semi-Axes of the Ellipse** The given equation of the ellipse is: \frac{x^2}{16} + \frac{y^2}{9} = 1 \implies \frac{x^2}{4^2} + \frac{y^2}{3^2} = 1 Here, the semi-major axis is a = 4 and the semi-minor axis is b = 3. **Step 2 — Set up the Definite Integral** Because the ellipse is symmetric with respect to both coordinate axes, the total area is 4 times the area in the first quadrant: y = \frac{3}{4}\sqrt{16 - x^2} A = 4 \int\_{0}^{4} \frac{3}{4}\sqrt{16 - x^2} \, dx = 3 \int\_{0}^{4} \sqrt{4^2 - x^2} \, dx **Step 3 — Evaluate the Definite Integral** Using the standard integration formula: A = 3 \left[\frac{x}{2}\sqrt{16 - x^2} + \frac{16}{2}\arcsin\left(\frac{x}{4}\right) \right]\_{0}^{4} A = 3 \left[\left(0 + 8\arcsin(1)\right) - \left(0 + 8\arcsin(0)\right) \right] A = 3 \left[8 \cdot \frac{\pi}{2} - 0 \right] = 3 \cdot 4\pi = 12\pi ━━━━━━━━━━━━━━━━━━━━━━━━━━━ ✅ **FINAL ANSWER:** **12\pi\text{ sq. units}** ━━━━━━━━━━━━━━━━━━━━━━━━━━━ 🎯 **KEY CONCEPTS:** - **Symmetry in Integration:** Multiplying the area of a single quadrant by 4 simplifies the limits of integration from [0, a] instead of [-a, a]. ⚠ **COMMON MISTAKES:** - **❌ Incorrect Approach:** Integrating without multiplying by 4, which yields only one quadrant of the ellipse (3\pi). - **✅ Correct Approach:** Multiply the first quadrant integral by 4 to find the entire bounded region. * * * ### Exercise 3: Area Bounded by the Circle in the First Quadrant ## Circle Area in the First Quadrant 🧠 **STEP-BY-STEP SOLUTION & OPTION ANALYSIS:** The circle equation is x^2 + y^2 = 4 = 2^2, which has its center at (0, 0) and radius r = 2. The bounded region is in the first quadrant, between x = 0 (the y-axis) and x = 2 (the rightmost boundary of the circle). - **Method 1 (Integration):** - **Method 2 (Geometric Formula):** The first quadrant represents exactly one-quarter of the total area of the circle: **Evaluation of Options:** - **(A) \pi:** Correct. Represents one quadrant of a circle of radius r = 2. - **(B) \frac{\pi}{2}:** Incorrect. Corresponds to an area of radius r = \sqrt{2}. - **(C) \frac{\pi}{3}:** Incorrect. Arithmetic error in evaluating the area. - **(D) \frac{\pi}{4}:** Incorrect. Corresponds to a unit circle quadrant (r = 1). ━━━━━━━━━━━━━━━━━━━━━━━━━━━ ✅ **FINAL ANSWER:** **(A) — \pi** ━━━━━━━━━━━━━━━━━━━━━━━━━━━ * * * ### Exercise 4: Area Bounded by the Parabola and the y-Axis ## Parabolic Region Bounded by the y-Axis 🧠 **STEP-BY-STEP SOLUTION & OPTION ANALYSIS:** The bounding curve is given by y^2 = 4x, along with the line y = 3 and the y-axis (x = 0). Since the region is bounded by the y-axis, it is most direct to integrate with respect to y from y = 0 to y = 3: x = \frac{y^2}{4} - **Step-by-Step Calculation:** - **Simplifying the Fraction:** **Evaluation of Options:** - **(A) 2:** Incorrect. - **(B) \frac{9}{4}:** Correct. Matches \int\_{0}^{3} \frac{y^2}{4} \, dy = \frac{9}{4}. - **(C) \frac{9}{3}:** Incorrect. Arises if the factor of 4 in the denominator is omitted or misapplied. - **(D) \frac{9}{2}:** Incorrect. Forgets the division by 3 during the antiderivative calculation \int y^2 dy = \frac{y^3}{3}. ━━━━━━━━━━━━━━━━━━━━━━━━━━━ ✅ **FINAL ANSWER:** **(B) — \frac{9}{4}** ━━━━━━━━━━━━━━━━━━━━━━━━━━━ Would you like to solve a similar practice problem involving integration along the y-axis?