Let a>0 and b>0. Show that the area of the ellipse (x^2)/(a^2)+(y^2)/(b^2)=1 is πa b (see figure

Let a>0 and b>0. Show that the area of the ellipse...

Key Concepts

Ellipse Geometry

An ellipse is a conic section defined by the equation x²/a² + y²/b² = 1, where a and b represent the semi-major and semi-minor axes respectively. Its shape is characterized by these two parameters and the standard formula for its area is ?ab, which shows the direct dependency of the area on the lengths of its axes.

Coordinate Transformation

To compute the area of the ellipse, a common technique is to perform a coordinate transformation that scales the axes. By substituting x = au and y = bv, the ellipse equation transforms into u² + v² = 1, which is the equation of a unit circle. This transformation simplifies the integration process by converting the problem into one of finding the area of a circle.

Jacobian Determinant

When performing a coordinate transformation in integration, the Jacobian determinant arises as a multiplicative factor that adjusts for the change in the variable domain. In the case of the ellipse transformation, the Jacobian of the transformation from (x, y) to (u, v) gives the factor ab. This ensures that after integrating over the unit circle (with area ?), the result is correctly scaled back to yield the area ?ab for the original ellipse.

Transcript

00:01 In this question, we have an ellipse whose equation is x square upon a square plus y square upon b squared equals to 1.

00:13 Here, a is greater than 0 and b is greater than 0.

00:20 We are required to prove that the area of this ellipse is pi a, b.

00:29 So let's see how to solve this question.

00:31 First of all, let's find the value of y from the equation of ellipse.

00:39 So we can write x square upon a square plus y square upon b square is equal to 1.

00:46 So we will have y square is equal to b square into 1 minus x square upon a square.

00:55 So, y will be equals to b upon a under root of a square minus x square.

01:09 And now let's write the equation to calculate the area of the ellipse.

01:16 So this will be equal to area a is equals to 4 into integration 0 to a, b upon a under root of a square minus x square d x and we can write it as 4b upon a integration 0 to a a square minus x square to the power 1 upon 2 dx.

01:57 And now to solve this integration, consider x is equals to a sine theta.

02:10 Therefore, by the differentiation we can write d x is equal to a cosine theta d theta.

02:21 And when we substitute x is equal to 0, we get theta is equal to 0.

02:27 And when we substitute x is equal to a, we get theta is equals to pi by 2.

02:36 Now substitute all the values in the above expression.

02:40 So we can write area a is equal to 4b upon a integration 0 to pi by 2 under root of a square minus a square sine theta into a cosine theta d theta...

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