Find the volume of the region under $z=x^2+y^2$ over $R$ bounded by $x=y^2$ with double integral?

Question

I couldn't figure out the limits of the double integral.This is my first question here so forgive me about any mistake and feel free to tell my mistakes to me.$$\int\int_Rx^2+y^2dA$$ In the answer key result is $$\int_{-1}^1\int_{y^2}^{2-y^2}(x^2+y^2)dxdy$$

Answer 1

The bounding surfaces are: $$x= y^2,\qquad x = 2 - y^2,\qquad z = x^2 + y^2,\qquad z = 0.$$ The triple integral is: $$\int_{-1}^1\int_{y^2}^{2 - y^2}\int_0^{x^2 + y^2}\,dzdxdy.$$ Why $[-1,1]$? Check the intersection points of $x = y^2$ and $x = 2 - y^2$.