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

[gigi9]

Someone please 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

 

[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?

 

[Hurkyl]

***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 please

 

[HallsofIvy]

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 πa², the semi-circle has area πa²/2 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]

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

 

[HallsofIvy]

Go back and read problem 1 again. Even though you typed it in here, apparently you did not understand what it was asking you to do.
You are, specifically, to use the formula for area of a circle to find the integral- NOT to "integrate the 1st one" in the usual sense.

As I said before, once you have found that integral, rewrite
x^2/a^2+ y^2/b^2= 1 as y= b√(1- x^2/a^2)= (b/a)√(a^2- x^2) (this is the top half of the ellipse) and use the integral in the first problem (or simply the formula for area of a circle) to find the area of an ellipse.