Find the area of the region bounded by the ellipse x²/16 + y²/9 = 1

Maths Question
1

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 .
Find the area of the region bounded by the ellipse x²/16 + y²/9 = 1 .1024×577 55.6 KB

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Solutions to Calculus Exercises: Areas Bounded by Curves

Exercise 1: Area of the Region Bounded by the Ellipse

:light_bulb: 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

:brain: 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

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:white_check_mark: FINAL ANSWER:
12\pi\text{ sq. units}
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:bullseye: 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].

:warning: COMMON MISTAKES:

  • :cross_mark: Incorrect Approach: Integrating without multiplying by 4, which yields only one quadrant of the ellipse (3\pi).
  • :white_check_mark: 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

:brain: 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):

    A = \int_{0}^{2} y \, dx = \int_{0}^{2} \sqrt{4 - x^2} \, dx
    A = \left[ \frac{x}{2}\sqrt{4 - x^2} + \frac{4}{2}\arcsin\left(\frac{x}{2}\right) \right]_{0}^{2} = 2 \arcsin(1) - 0 = 2 \cdot \frac{\pi}{2} = \pi

  • Method 2 (Geometric Formula):
    The first quadrant represents exactly one-quarter of the total area of the circle:

    A = \frac{1}{4}(\pi r^2) = \frac{1}{4}(\pi \cdot 2^2) = \pi

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).

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:white_check_mark: FINAL ANSWER:
(A) — \pi
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Exercise 4: Area Bounded by the Parabola and the y-Axis

Parabolic Region Bounded by the y-Axis

:brain: 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:

    A = \int_{0}^{3} x \, dy = \int_{0}^{3} \frac{y^2}{4} \, dy
    A = \frac{1}{4} \left[ \frac{y^3}{3} \right]_{0}^{3} = \frac{1}{12} \left[ 3^3 - 0 \right] = \frac{27}{12}

  • Simplifying the Fraction:

    \frac{27}{12} = \frac{9}{4}

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}.

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:white_check_mark: FINAL ANSWER:
(B) — \frac{9}{4}
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Would you like to solve a similar practice problem involving integration along the y-axis?