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Find the area bounded by the curves $y=-\sqrt{-x}$ and $x=-\sqrt{-y}$,where $x,y\leq 0$.

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I found the area $A=\int_{-1}^{0}-x^2+\sqrt{-x} dx=-1$

when we take its absolute value,it becomes $+1$ but the answer given in my book is $\frac{1}{3}$.Have i done some mistake?What is the right way to solve it?

diya
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  • $\int_{-1}^{0}-x^2+\sqrt{-x} dx=\frac13$. Correct reasoning, error in the computation. – imj Sep 27 '15 at 11:52

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The formula is correct: $\int_{-1}^{0}-x^2+\sqrt{-x} dx$. It is easier to compute if you take the symetry with respect to the origin, ie the area between $y=x^2$ and $y=\sqrt{x}$ with $x\in[0;1]$ :

\begin{align*} A=\int_{-1}^{0}-x^2+\sqrt{-x} dx &= \int_{0}^{1}\sqrt{x}-x^2 dx\\ &=\left[\dfrac23x^{\frac32}-\dfrac{x^3}3\right]_0^1\\ &= \left(\dfrac23-\dfrac13 \right)-\left(0-0 \right)\\ &=\dfrac13 \end{align*}

imj
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    @diya : Of course it can. Its just that the antiderivative of $x\mapsto \sqrt{x}$ is well-known, whereas the antiderivative of $x\mapsto \sqrt{-x}$ requires thinking. – imj Sep 27 '15 at 12:15