# Use the known area of a circle to find the value of the integral

gigi9
Someone plz show me how to do the problem below, thanks very much.
1)***Use the known area of a circle to find the value of the integral
integral from -a to a of the function sqrt(a^2-x^2)dx.
2)***Then use the result of this integral to find the enclosed area of (x^2)/(a^2)+(y^2)/(b^2)= 1, a>b>0.
Plz show me how to integrate #1

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## Answers and Replies

StephenPrivitera
Use the substitution x=asinu so that sqrt(a2-x2)=acosu.

StephenPrivitera
Originally posted by gigi9
***Then use the result of this integral to find the enclosed area of (x^2)/(a^2)+(y^2)/(b^2), a>b>0.

The enclosed area of what?

Staff Emeritus
Gold Member
***Use the known area of a circle to find the value of the integral
integral from -a to a of the function sqrt(a^2-x^2)dx. ***

You know that the definite integral of a positive function is the area between its graph and the x-axis right? What is the graph of sqrt(a^2-x^2)?

gigi9
still confused..explain more plz

Homework Helper
You seem to have serious problems with basic concepts- as illustrated by your saying "Then use the result of this integral to find the enclosed area of (x^2)/(a^2)+(y^2)/(b^2), a>b>0."
I presume that you copied this from some problem but you even copied wrong. "(x^2)/(a^2)+(y^2)/(b^2)" does not enclose anything- it is not a graph nor a function nor an equation. I suspect that you book had "(x^2)/(a^2)+(y^2)/(b^2)= 1", the equation of an ellipse.

As for the first problem: If you are expected to be able to do integrals, then you should already know that a basic interpretation of "integral" is "area under a curve". The function y= sqrt(a^2-x^2)dx is the upper half of the circle x^2+ y^2= a^2 (you can see that by squaring both sides of the given equation). Since the circle has area &pi;a2, the semi-circle has area &pi;a2/2 and that is the value of the integral of the function.

Now that you know that integral, solve (x^2)/(a^2)+(y^2)/(b^2)= 1 for y and apply that knowledge.

gigi9
plz show me how to integrate the 1st one...and how to find the enclosed area of the second one plz...(maybe the first few step or something to get me started...) Thanks a lot.