Sketch the region R bounded by y = x^2, y = 2 - x, x = 0 and x = 1, then evaluate the double integral of x^2 + y^2 over R with respect to y then x.

Step-by-step solution

1. 1

   Describe the region before writing any limits. On the strip $0\le x\le 1$ compare the two curves. At $x=0$ they give $y=0$ and $y=2$; at $x=1$ they give $y=1$ and $y=1$. Since $2-x-x^2=(2+x)(1-x)\ge 0$ on $[0,1]$, the parabola is the lower boundary and the line is the upper one throughout, meeting only at the corner $(1,1)$.

2. 2

   Set up the iterated integral in the order dy dx. Because the vertical slice at each $x$ runs cleanly from one curve to the other, this order needs no splitting:

   $I=\int_{0}^{1}\int_{x^{2}}^{2-x}\left(x^{2}+y^{2}\right)dy\,dx.$

3. 3

   Do the inner integral, treating x as a constant.

   $\int_{x^{2}}^{2-x}\left(x^{2}+y^{2}\right)dy=\left[x^{2}y+\frac{y^{3}}{3}\right]_{y=x^{2}}^{y=2-x}$

   $=x^{2}\left(2-x-x^{2}\right)+\frac{(2-x)^{3}-x^{6}}{3}.$

4. 4

   Expand into a single polynomial in x. Using $(2-x)^3=8-12x+6x^2-x^3$,

   $=2x^{2}-x^{3}-x^{4}+\frac{8-12x+6x^{2}-x^{3}-x^{6}}{3}=\frac{8}{3}-4x+4x^{2}-\frac{4x^{3}}{3}-x^{4}-\frac{x^{6}}{3}.$

5. 5

   Integrate term by term from 0 to 1.

   $I=\frac{8}{3}-2+\frac{4}{3}-\frac{1}{3}-\frac{1}{5}-\frac{1}{21}.$

   Over the common denominator $105$ this is

   $I=\frac{280-210+140-35-21-5}{105}=\frac{149}{105}.$

6. 6

   Check the size of the answer. The region has area $\int_0^1(2-x-x^2)dx=2-\tfrac12-\tfrac13=\tfrac76\approx1.167$, and $x^2+y^2$ ranges from $0$ up to $4$ on it, so an average value near $\tfrac{149/105}{7/6}\approx1.22$ is entirely plausible. A Monte-Carlo estimate over the same region returns $1.4195$, matching $149/105\approx1.4190$.