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What is the area included in the curves

√x + √|y| = 1 and |x| + |y| = 1.

I know the area the total area of the |x| + |y| = 1 is 2 units square but i cannot determine the area of the given first curve .. Please help.

Yash Gupta
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    I think you can use symmetry and only analyse the first quadrant ($x>0$ and $y>0$) – Matti P. Jul 03 '19 at 10:22
  • @PeterForeman Area common in between these two curves – Yash Gupta Jul 03 '19 at 10:24
  • @PeterForeman Can we calculate the total area like calculate in first quadrant and rest of them by symmetry ? – Yash Gupta Jul 03 '19 at 10:26
  • @MattiP. Sorry but can you help me how to calculate the area in the first quadrant in these curves ? – Yash Gupta Jul 03 '19 at 10:28
  • For the second one, I would use symmetry. If we only consider $x,y>0$ then the equation reduces to $x+y=1$ or $y=1-x$. This is just a triangle with side lengths $1\times 1$, giving an area of $\frac{1}{2}\times 1 \times 1$. So going back, symmetry gives us four of these, making the total area $4 \times \frac{1}{2}= 2$. – Matti P. Jul 03 '19 at 10:34
  • @MattiP. Okay but what about the first equation with the √|y| ? – Yash Gupta Jul 03 '19 at 10:35
  • here is the graph, however, it is not clear which included area is expected. – farruhota Jul 03 '19 at 11:03

2 Answers2

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Hint: Since $$x\geq 0$$ we get $$\sqrt{x}+\sqrt{|y|}=1$$ and $$x+|y|=1$$ and you can compute $y$

Dr. Sonnhard Graubner
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The branch of $\sqrt{|x|}+\sqrt{|y|}=1$ in the first quadrant is $$ y=(1-\sqrt{x})^2=1-2\sqrt{x}+x $$ The area under it is $$ \int_0^1(1-2\sqrt{x}+x)\,dx $$ The area under the line $x+y=1$ (from $x=0$ to $x=1$) is…

Subtract and multiply by $4$.

egreg
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