Multiple integrals: fiind the region bounded by $y=2-x, y=0$ and $x=4-y^2$ in the first quadrant

Question

Sketch the regions of integration and evaluate the integral. Can someone help me set this up I know the $y$ bounds but would the lower $x$ bound just be $0$?

Answer 1

Setting up the bounds of the integral is made easier once you draw the region.

You have

$$\iint_R 3xy\,\mathrm{d}A=3\int_0^2\int_{2-y}^{4-y^2}xy\,\mathrm{d}x\,\mathrm{d}y$$

Alternatively, you can compute the integral in the opposite direction, such that

$$\iint_R 3xy\,\mathrm{d}A=3\left(\int_0^2\int_{2-x}^\sqrt{4-x}xy\,\mathrm{d}y\,\mathrm{d}x+\int_2^4\int_0^\sqrt{4-x}xy\,\mathrm{d}y\,\mathrm{d}x\right)$$ where in this case, you need to split up the region of integration into two segments because the range of the $y$ coordinate of each point in the region isn't uniformly described by one lower bound.

Another option would be to find a different change of coordinates, but unless you're explicitly told to do so, the first method should suffice.