/SCI! help me with this double integral!

https://www.symbolab.com/solver/double-integrals-calculator

>>9003286
>The answer is trivial

>>9003286
I get it to be 8/3.
What are you struggling with?

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Let x=0. Then 0 <=y <=4.

Moreover 0 <=x <=(2-(y)^1/2)^2

Integrate over these domains.

>>9003286
[math]\sqrt{x}+\sqrt{y}=2[/math]
[math]x+y=4[/math]
[math]y=4-x[/math]
[math]\int{(4-x)} dx=4x-\frac{x^2}{2}[/math]

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>>9004203
Are you an actual retarded

>>9004216
Oh sorrry, I forgot a step.
[math]4(2)-\frac{2^2}{2}[/math]
[math]8-2[/math]
[math]6*2[/math]
[math]12[/math]

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>>9004230

Do this, keeping in mind both [math]x[/math] and [math]y[/math] are non-negative:
[eqn]\sqrt x \,+\, \sqrt y \,=\, 2 \quad\text{then}\quad y \,=\, \left( 2 \,-\, \sqrt x \right)^2 \,=\, 4 \,-\, 4\,\sqrt x \,+\, \left( \sqrt x \right)^2.[/eqn]
Therefore,
[eqn]\mathscr A \,=\, \int_0^4 \left( 4 \,-\, 4\,\sqrt x \,+\, x \right)\, \mathrm dx \,=\, 4 \,\times\, 4 \,-\, 4 \,\left( \frac{2\, \sqrt{4^3}}3 \,-\, \frac{2\, \sqrt{0^3}}3 \right) \,+\, \left( \frac{4^2}2 \,-\, \frac{0^2}2 \right) \,=\, 16 \,-\, 4 \, \frac{2\,\sqrt{8^2}}3 \,+\, \frac{16}2 \,=\, 16 \,-\, \frac{8 \,\times\, 8}3 \,+\, 8 \,=\, 24 \,-\, \frac{64}3 \,=\, \frac{72 \,-\, 64}3 \,=\, \frac83[/eqn]

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>>9004268
I think OP wants it framed as a double integral

See
>>9004185

Perfect otherwise

Why are you faggots helping this underage b8 with his homework?