Question 1 of 9intermediate🔧 ApplyNumerical4 marks

Find the area of the region bounded by the ellipse x216+y29=1\frac{x^2}{16} + \frac{y^2}{9} = 1.

Correct Answer

The area of the region bounded by the ellipse is 12π12\pi square units.

Exercise: EXERCISE 8.1 | Q: 1 | (Page Page 5)
For More Understanding

Explanation

This question is from Exercise 8.1, Question 1. The solution follows Example 2 from the textbook which derives the general formula for area of an ellipse as πab. Students need to identify the values of a and b from the given equation and apply the integration method shown in the context.

Solution Steps

  1. Step 1: Compare the given ellipse equation x216+y29=1\frac{x^2}{16} + \frac{y^2}{9} = 1 with the standard form x2a2+y2b2=1\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1. We get a2=16⇒a=4a^2 = 16 \Rightarrow a = 4 and b2=9⇒b=3b^2 = 9 \Rightarrow b = 3.

  2. Step 2: Since the ellipse is symmetrical about both axes, the total area = 4∫0ay dx4 \int_{0}^{a} y \, dx (taking vertical strips in the first quadrant).

  3. Step 3: From the ellipse equation, solve for y: y=±baa2−x2y = \pm \frac{b}{a} \sqrt{a^2 - x^2}. Since the region lies in the first quadrant, y is positive: y=baa2−x2y = \frac{b}{a} \sqrt{a^2 - x^2}.

  4. Step 4: Substitute values and integrate: Area = 4∫043416−x2 dx=3[x216−x2+162sin⁡−1x4]044 \int_{0}^{4} \frac{3}{4} \sqrt{16 - x^2} \, dx = 3 \left[ \frac{x}{2} \sqrt{16 - x^2} + \frac{16}{2} \sin^{-1}\frac{x}{4} \right]_{0}^{4}.

  5. Step 5: Evaluate at limits: At x = 4: 42×0+8sin⁡−1(1)=8×π2=4π\frac{4}{2} \times 0 + 8 \sin^{-1}(1) = 8 \times \frac{\pi}{2} = 4\pi. At x = 0: 0. Therefore, Area = 3×4π=12π3 \times 4\pi = 12\pi square units.

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