How to manually sketch the graph of $\sqrt x+\sqrt y=1$ in order to find the area bounded by the curves $x+y=1$ and $\sqrt x+\sqrt y=1$ ?
The graph of the first function doesn't seem to any standard graph while the second one is the equation of a straight line . Any suggestions on how to roughly sketch it so that I can find the required area ?
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Take the graph of $x+y=1,$ which is a line thru $(0,1)$ and $(1,0).$ Take a few points $(x,y)$ on this line with $x\in [0,1/2].$ For each of them the points $(x^2,y^2)$ and $(y^2,x^2)$ will lie on the graph you wish to sketch. It won't take many points to get a good idea of the shape.
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To get an idea on how to sketch the graph, we can implicitly differentiate $\sqrt x+\sqrt y = 1$ to get $y' = -\sqrt\frac y x$ and $y'' = \frac {\sqrt x+\sqrt y} {2x^{\frac 3 2}}$ (note that both $x$ and $y$ fall inside the closed interval $[0,1]$). So, the graph will be a decreasing function from $0$ to $1$ (since $y'\leqslant0\forall x,y$) that is concave up ($y''\geqslant0\forall x,y$). That is as far as we can go regarding sketching the graph.
As for finding the area between the two curves, since the whole graph is above the $x$ axis, we can rewrite the equation as $y = 1+x-2\sqrt x, x\in[0,1]$. So the area bound between the two curves will be (note that the area bounded by $x+y=1$ above the $y$ axis between $[0,1]$ is the area of a perpendicular triangle of side lengths $1$, so its area is $\frac 1 2$)
$\displaystyle\frac 1 2-\int_0^11+x-2\sqrt xdx=\frac 1 2 -\frac 1 6=\frac 1 3$
You don't require to sketch the graph to find the area. Find where the teri graphs intersect. $√x+√y=1$ $x+y+2√xy=1$ But, $x+y=1$
So$√xy=0$
$x=0$. Or. $y=0$.
So when $x=0,y=1$ $y=0,x=1$
$(0,1). (1,0)$ are where they intersect.
Now you can use integration to find the area bounded by the curve and line between points $(1,0) ; (0,1)$ on X axis.
And the difference will give you the area between them.