$\iint_V |y-x^{2}| \operatorname{d}x \operatorname{d}y$ with $V = [-1,1] \times [0,2]$

Question

it's especially difficult because i don't understand how to integrate absolute value terms. I only know that if you function, say $x^{2}-1$, is below the $x$-axis i need to integrate $1-x^2$ between the interval $[-1,1]$.

But in this case i got two variables which quite confuse me....

Can someone give me tips/hints/help with this problem?

Thanks!

Answer 1

For every $x$ in $[-1,1]$, $$I(x)=\int_0^2|y-x^2|\mathrm dy=\int_0^{x^2}(x^2-y)\mathrm dy+\int_{x^2}^2(y-x^2)\mathrm dy$$ hence $$ I(x)=\left[x^2y-\tfrac12y^2\right]_0^{x^2}+\left[\tfrac12y^2-x^2y\right]_{x^2}^2=\ldots$$

Answer 2

Anyway you have to take some method to remove the absolute value symbol. The simplest way is to divide the region of integration into parts according to the sign of the value in the absolute value symbol. Indeed it will make the boundaries more complex. But I don't think it is evitable.