How to find the area between $y=x^2-8$ and $y=-x^2$

Key concept: This is a classic “area between curves” problem. The key move is to first find where the parabolas meet, then subtract the lower curve from the upper one over that interval.

Step by step

Step 1: Find the intersection points

Set the curves equal:

$x^2-8=-x^2$

$2x^2=8$

$x^2=4 \Rightarrow x=\pm 2$

Step 2: Identify the top curve

Between $x=-2$ and $x=2$,

• upper curve: $y=-x^2$
• lower curve: $y=x^2-8$

So the vertical distance is

$(-x^2)-(x^2-8)=8-2x^2$

Step 3: Integrate the difference

$A=\int_{-2}^{2}(8-2x^2)\,dx$

Because the integrand is even:

$A=2\int_{0}^{2}(8-2x^2)\,dx$

$=2\left[8x-\frac{2x^3}{3}\right]_{0}^{2}$

$=2\left(16-\frac{16}{3}\right)=2\cdot \frac{32}{3}=\frac{64}{3}$

Answer

$\boxed{\frac{64}{3}}$

Pitfall alert

A common slip is reversing the curves and integrating a negative difference. If your setup gives a negative value, your top-minus-bottom order is backwards. Also, don’t forget there are two intersection points, not just one.

Try different conditions

If the curves were shifted so they still intersect twice but not symmetrically about the $y$-axis, you’d use the same method: solve for the new bounds, then integrate top minus bottom on that interval. The only thing that changes is the algebra in the subtraction and the limits of integration.

Further reading

area between curves, definite integral, intersection points