Find the area of the region bounded by the ellipse \(\frac{x^2}{16}+\frac{y^2}{9}=1\)
Find the area of the region bounded by the ellipse \(\frac{x^2}{16}+\frac{y^2}{9}=1\)
Solution and Explanation
The given equation of the ellipse, \(\frac{x^2}{16}+\frac{y^2}{9}=1\) ,can be represented as
It can be observed that the ellipse is symmetrical about x-axis and y-axis.
∴Area bounded by ellipse=4×Area of OAB
Area of OAB=\(\int^4_0ydx\)
=\(\int^4_03\sqrt{1-\frac{x^2}{16}}dx\)
=\(\frac{3}{4}\int^4_0\sqrt{16-x^2}dx\)
=\(\frac{3}{4}\bigg[\frac{x}{2}\sqrt{16-x^2}+\frac{16}{2}\sin^{-1}\frac{x}{4}\bigg]^4_0\)
=\(\frac{3}{4}\)[2\(\sqrt{16-16}\)+8sin-1(1)-0-8sin-1(0)]
=\(\frac{3}{4}\bigg[\frac{8\pi}{2}\bigg]\)
=\(\frac{3}{4}[4\pi]\)
=3\(\pi\)
Therefore, area bounded by the ellipse= 4×3\(\pi\)=12\(\pi\) units.
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Concepts Used:
Integral
The representation of the area of a region under a curve is called to be as integral. The actual value of an integral can be acquired (approximately) by drawing rectangles.
- The definite integral of a function can be shown as the area of the region bounded by its graph of the given function between two points in the line.
- The area of a region is found by splitting it into thin vertical rectangles and applying the lower and the upper limits, the area of the region is summarized.
- An integral of a function over an interval on which the integral is described.
Also, F(x) is known to be a Newton-Leibnitz integral or antiderivative or primitive of a function f(x) on an interval I.
F'(x) = f(x)
For every value of x = I.
Types of Integrals:
Integral calculus helps to resolve two major types of problems:
- The problem of getting a function if its derivative is given.
- The problem of getting the area bounded by the graph of a function under given situations.