Why does the integral of an ellipse equation only find a fourth of an ellipse's area.

Integrate

Integrate

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I am getting it from this text book writing.

Ellipse integral.png

I get that it's symmetric to on both axes but I don't know how to put it all together in my head.

Sorry for the dumb question. Thank you.
 
Hi.
Do you understand why the author chose the positive expression for y? Why the interval of integration is between 0 and a?
 
Look at the given diagram. Is it not clear to you that the blue area is one-fourth of the area of the ellipse?
Does \(\displaystyle\int_0^a {\frac{b}{a}\sqrt {{a^2} - {x^2}} dx} \) give the blue area?
 
There is no integral of an ellipse equation.

You could have integrated from -a to a and gotten 1/2 the area of the ellipse.

You need to understand what the limits of an integral mean and what the integrand means. When you learn that then and only then will you understand the answer to your question.
 
The first problem is that the equation of an ellipse does not describe a function. Remember the vertical line test from 9th grade? And we integrate functions. If we go

[MATH]\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1 \implies y^2 = b^2 - \dfrac{b^2x^2}{a^2} = \dfrac{a^2b^2}{a^2} - \dfrac{b^2x^2}{a^2} \implies \\ y^2 = \dfrac{b^2}{a^2} * (a^2 - x^2 ) \implies\\ y = \text {PLUS or MINUS } \dfrac{b}{a} * \sqrt{a^2 - x^2}.[/MATH]You cannot integrate that because it is not a function; insert a value for x, and you get two possible values for y. So you have to pick one or the other as the function to integrate. One will give you an area above the x-axis. The other will give you an area below the x-axis.

Now, whichever area you choose, you can set the limits of integration from - a to + a. And multiply by 2. Or you can set the limits of integration from 0 to a and multiply by 4. Which do you think is easier?
 
pka said:
Look at the given diagram. Is it not clear to you that the blue area is one-fourth of the area of the ellipse?
Does \(\displaystyle\int_0^a {\frac{b}{a}\sqrt {{a^2} - {x^2}} dx} \) give the blue area?
Click to expand...
Yeah I see that but I'm having trouble understanding why it's not the entire ellipse. It feels super close to me I just can't wrap my mind around it.
 
Jomo said:
There is no integral of an ellipse equation.

You could have integrated from -a to a and gotten 1/2 the area of the ellipse.

You need to understand what the limits of an integral mean and what the integrand means. When you learn that then and only then will you understand the answer to your question.
Click to expand...
So why did the author choose 0 to a?
 
JeffM said:
The first problem is that the equation of an ellipse does not describe a function. Remember the vertical line test from 9th grade? And we integrate functions. If we go

[MATH]\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1 \implies y^2 = b^2 - \dfrac{b^2x^2}{a^2} = \dfrac{a^2b^2}{a^2} - \dfrac{b^2x^2}{a^2} \implies \\ y^2 = \dfrac{b^2}{a^2} * (a^2 - x^2 ) \implies\\ y = \text {PLUS or MINUS } \dfrac{b}{a} * \sqrt{a^2 - x^2}.[/MATH]You cannot integrate that because it is not a function; insert a value for x, and you get two possible values for y. So you have to pick one or the other as the function to integrate. One will give you an area above the x-axis. The other will give you an area below the x-axis.

Now, whichever area you choose, you can set the limits of integration from - a to + a. And multiply by 2. Or you can set the limits of integration from 0 to a and multiply by 4. Which do you think is easier?
Click to expand...
Thank you thank you thank you. This made it abundantly clear. Complete light bulb. Thank you.
 
Integrate said:
So why did the author choose 0 to a?
Click to expand...
Because evaluating the integral at x=0 is usually easier than evaluating the integral at x=-a.
As I said you can choose between integrating from -a to a or from 0 to a. Most people choose from 0 to a over -a to a.
 
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