- Thread starter akleron
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So it should be pretty clear that the outer integration limits, for \(\displaystyle y\) are \(\displaystyle 0 \to 1\)

Now look at your graph. As you move \(\displaystyle y\) in the positive direction, as a function of \(\displaystyle y\)

what value of \(\displaystyle x\) does your area start at? What value does it end at?

Those will be your lower and upper integration limits for \(\displaystyle x\)

so the integration limits for x should be from y=x^2 to

So it should be pretty clear that the outer integration limits, for \(\displaystyle y\) are \(\displaystyle 0 \to 1\)

Now look at your graph. As you move \(\displaystyle y\) in the positive direction, as a function of \(\displaystyle y\)

what value of \(\displaystyle x\) does your area start at? What value does it end at?

Those will be your lower and upper integration limits for \(\displaystyle x\)

y=√x ?

No.so the integration limits for x should be from y=x^2 to

y=√x ?

The area starts at \(\displaystyle y = \sqrt{x} \Rightarrow x = y^2\)

It ends at \(\displaystyle y = x^2 \Rightarrow x = \sqrt{y}\)

Thus the limits are \(\displaystyle x: y^2 \to \sqrt{y}\)

so the overall area is

\(\displaystyle A = \displaystyle \int_0^1 \int_{y^2}^{\sqrt{y}}~dx~dy\)

yeah, you got it right in the next post. Cheers.

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