Find the area of the region bounded by and above the x-axis.

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- Dec 29th 2008, 06:33 AMvarunnayuduBounded region.
Find the area of the region bounded by and above the x-axis.

- Dec 29th 2008, 07:00 AMLast_Singularity
Do you actually need to carry out the integration?

If not, you can simply use the fact that the area of an ellipse is . And since you are only looking for the area above the x-axis, you just cut that in half to get: - Dec 29th 2008, 07:04 AMIsomorphism
Observe that

with the substitution ,

Now you can continue :D

**Aliter:**

Define and . Now we can see that

transforms

Thus (Area of the semi circle) - Dec 29th 2008, 07:13 AMSoroban
Hello, varunnayudu!

This is a classic problem -- the area of an ellipse (well, semi-ellipse).

Quote:

Find the area of the region bounded by and above the x-axis.

The graph is an ellipse, center at (0,0), major axis (ħa,0), minor axis (0, ħb)

Due its symmetry, we can find the area in Quadrant 1 and double.

Solving for

The area of the semi-ellipse is: .

Let:

Substitute: .

. . . . . .

Back-substitute: .

. . And we have: .

Evaluate: .

- Dec 29th 2008, 07:19 AMvarunnayuducan u explain me what u have done
Can u explain me what u have done here

with the substitution http://www.mathhelpforum.com/math-he...8458655f-1.gif,

http://www.mathhelpforum.com/math-he...a0dac5f7-1.gif

how did u get - Dec 29th 2008, 07:38 AMIsomorphism
Perhaps my post was not clear.

Soroban's post explains this step clearly. Let: . Now substitute and in the integral. Since x is varying from -a to a, will vary from to . Substitute the limit also.

Now do you see what i have done?

This idea is called integration by substitution. Haven't you seen it before?