Find the area of the region bounded by the ellipse \(\frac{x^2}{16}+\frac{y^2}{9}=1\)
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.
What is the Planning Process?
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.
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.
Integral calculus helps to resolve two major types of problems: