Can I have some help on this integration question? Watch

fuzzybear
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Find the area under the curve (x^2 / a^2) + (y^2 / b^2) = 1 in the first quadrant, i.e. in the positive X and Y axis

I tried writing out the curve's equation with y the subject:

y^2 = b^2 - (b^2)(x^2)/a

but I don't know how I should integrate it from here. And also, how do I know what the limits would be?
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Indeterminate
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(Original post by fuzzybear)
Find the area under the curve (x^2 / a^2) + (y^2 / b^2) = 1 in the first quadrant, i.e. in the positive X and Y axis

I tried writing out the curve's equation with y the subject:

y^2 = b^2 - (b^2)(x^2)/a

but I don't know how I should integrate it from here. And also, how do I know what the limits would be?
You could let

\displaystyle y = \sqrt{b^2 - \dfrac{(xb)^2}{a}} = \sqrt{b^2\left(1-\dfrac{x^2}{a}\right)}=b\sqrt{1-\dfrac{x^2}{a}}

which is an integral that can be done by substitution, say

x=a^{\frac{1}{2}} \sin(u)

This curve is actually an ellipse. You can determine the limits by finding the point where it crosses the x axis (in the first quadrant)
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fuzzybear
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(Original post by Indeterminate)
You could let

\displaystyle y = \sqrt{b^2 - \dfrac{(xb)^2}{a}} = \sqrt{b^2\left(1-\dfrac{x^2}{a}\right)}=b\sqrt{1-\dfrac{x^2}{a}}

which is an integral that can be done by substitution, say

x=a^{\frac{1}{2}} \sin(u)

This curve is actually an ellipse. You can determine the limits by finding the point where it crosses the x axis (in the first quadrant)
Thanks man

two questions:

- I substituted that, integrated it and got: b[sinu]

do I have to change sinu back in terms of x, or can I just sub in the equivalent u limits (to the x limits)?


- also, I've tried finding the point where it crosses the x-axis, by subbing in y = 0 for:

y = b*(1 - x^2 / a)^(1/2)

so that either b = 0, or (1 - x^2 / a)^(1/2) = 0

so would x = square root of a ?
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