double integral, how should i determine the integral boundaries ?

akleron

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Dec 28, 2019
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32
I need to calculate the following integral shown in this picture.
I think I managed to draw the area of D correctly [shown in the picture ] but i don't know how to continue..any help ?

1579207867469.png
 

Romsek

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\(\displaystyle x^2 = \sqrt{x} \Rightarrow x = 1\)

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 \(\displaystyle x\)
 

akleron

New member
Joined
Dec 28, 2019
Messages
32
\(\displaystyle x^2 = \sqrt{x} \Rightarrow x = 1\)

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
y=√x ?
 

akleron

New member
Joined
Dec 28, 2019
Messages
32
I think I got the idea now.
It's still hard for me to determine the integration boundaries but I guess that I will figure it out ...
Thanks for the help !
1579211699685.png
 

Romsek

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Joined
Nov 16, 2013
Messages
784
so the integration limits for x should be from y=x^2 to
y=√x ?
No.

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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