I think because y is a function of a?Hi.
Do you understand why the author chose the positive expression for y? Why the interval of integration is between 0 and a?
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.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?
So why did the author choose 0 to a?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.
Thank you thank you thank you. This made it abundantly clear. Complete light bulb. Thank you.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?
Not quite. a and b are fixed parameters. Try to follow JeffM reply.I think because y is a function of a?
Because evaluating the integral at x=0 is usually easier than evaluating the integral at x=-a.So why did the author choose 0 to a?
No, y is a function of xI think because y is a function of a?
Yes yes I think I get it now.No, y is a function of x