Evaluate the double integral of y over the region -1 <= x <= 1, x^2 <= y <= 1.

Step-by-step solution

1. 1

   Identify the region. For each $x$ in $[-1,1]$ the slice runs from the parabola $y=x^2$ up to the line $y=1$. The two curves meet where $x^2=1$, i.e. at $x=\pm 1$, so the region is exactly the area enclosed between the parabola and the horizontal line — closed, bounded, and symmetric about the $y$-axis.

2. 2

   Integrate in $y$ first, because the limits depend on $x$.

   $\int_{x^2}^{1} y\,dy=\left[\frac{y^2}{2}\right]_{y=x^2}^{y=1}=\frac{1}{2}-\frac{\left(x^2\right)^2}{2}=\frac{1-x^4}{2}$

   Note $\left(x^2\right)^2=x^4$, not $x^2$ — squaring the lower limit is the step that generates the quartic.

3. 3

   Use the even symmetry to halve the work. The function $\tfrac{1-x^4}{2}$ is even, so

   $\int_{-1}^{1}\frac{1-x^4}{2}\,dx=2\int_0^1\frac{1-x^4}{2}\,dx=\int_0^1\left(1-x^4\right)dx$

4. 4

   Finish the outer integral.

   $\int_0^1\left(1-x^4\right)dx=\left[x-\frac{x^5}{5}\right]_0^1=1-\frac15=\frac45$

5. 5

   Check against a centroid interpretation. The area of the region is $\int_{-1}^1\left(1-x^2\right)dx=2-\tfrac23=\tfrac43$. Since $\iint y\,dA=\bar y\cdot\text{Area}$, the answer implies a centroid height

   $\bar y=\frac{4/5}{4/3}=\frac{3}{5}=0.6$

   That is a sensible value: the region is widest near $y=1$ and pinches to a point at $y=0$, so its centre of mass should sit above the midpoint $y=0.5$ $\checkmark$.