Set up a definite integral for the area of the region lying in the right half of the ellipse given by x^2+9y^2=1

Question

I need help solving this. I have tried multiple times, and each time I get a different answer.

Answer 1

Since they gave you the problem in terms of x and y, it'd probably be best to solve it using x and y. Let's isolate y so we get a solid, "integratable" function.

x² + 9y² = 1 becomes y = ±√¯1/9¯(1 - x²)¯. If we take the positive square root, we are drawing the top half of the ellipse (type that into desmos.com for a good visual). Because the x² term has a coefficient of 1, we know the ellipse intersects the x-axis at (-1, 0) and (1, 0).

From there, we can integrate the square root expression from x=0 to x=1 to get the top right corner (or 1/4) of the ellipse:

∫ √¯1/9¯(1 - x²)¯ dx

0 to 1

Since that only represents the half of what the problem asks for, we multiply the integral by 2 to get the entire right hand side:

2 * ∫ √¯1/9¯(1 - x²)¯ dx

0 to 1

As you can see, it's a pretty nasty integral, so that's probably why the problem only asks for setup. If you wanted to find the answer, it'd be easier to use polar coordinates or simply the formula for an ellipse's area (you get π/6).

Answer 2

y = ± √(1 - x²)/3. Region what we have described by enequolities: - √(1 - x²)/3 ≤ y ≤ √(1 - x²)/3 and -1 ≤ x ≤ 1

So, A = 2/3∫₀¹√(1 -x²)dx