We will find this area by integrating with respect to y.
x= 8-8y^2 x= 8y^2 - 8
This integral is obtained hy taking the right hand function minus the left hand function or (8-8y^2 - ( )

To find the area between the curves \( x = 8 - 8y^2 \) and \( x = 8y^2 - 8 \) by integrating with respect to \( y \), we follow these steps:

### Step 1: Set up the Integral
The area \( A \) between the curves is given by the integral of the right-hand function minus the left-hand function over the range of \( y \), where the curves intersect.

Right-hand function: \( x = 8 - 8y^2 \)  
Left-hand function: \( x = 8y^2 - 8 \)

The integral for the area is:
\[
A = \int_{y_1}^{y_2} \left[ (8 - 8y^2) - (8y^2 - 8) \right] \, dy
\]

### Step 2: Simplify the Expression
Simplifying the expression inside the integral:

\[
(8 - 8y^2) - (8y^2 - 8) = 8 - 8y^2 - 8y^2 + 8 = 16 - 16y^2
\]

So, the area integral becomes:
\[
A = \int_{y_1}^{y_2} (16 - 16y^2) \, dy
\]

### Step 3: Find the Limits of Integration
The limits of integration \( y_1 \) and \( y_2 \) are the \( y \)-values where the curves intersect. To find these, set the two functions equal to each other:
\[
8 - 8y^2 = 8y^2 - 8
\]

Solving for \( y \):
\[
8 = 16y^2
\]
\[
y^2 = \frac{1}{2}
\]
\[
y = \pm \frac{1}{\sqrt{2}}
\]

So the limits of integration are \( y_1 = -\frac{1}{\sqrt{2}} \) and \( y_2 = \frac{1}{\sqrt{2}} \).

### Step 4: Perform the Integration
Now, we can calculate the integral:
\[
A = \int_{-\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}} (16 - 16y^2) \, dy
\]

We can now proceed with evaluating this integral to find the area.

Would you like to see the detailed steps for solving the integral or proceed directly to the result?

Here are some related questions you might consider:
1. What is the geometric interpretation of this area?
2. How would you compute the area if the curves were different?
3. What happens if we switch the limits of integration?
4. Can this be done by integrating with respect to \( x \)?
5. How would symmetry play a role in this integral?

**Tip**: Always check for symmetry in the region of integration to simplify calculations.

We will find this area by integrating with respect to y. x = 8 - 8y^2, x = 8y^2 - 8. This integral is obtained by taking the right-hand function minus the left-hand function or (8 - 8y^2 - (8y^2 - 8)).

Learn how to find the area between the curves x = 8 - 8y^2 and x = 8y^2 - 8 by integrating with respect to y. This problem involves setting up the integral, simplifying the expression, finding the limits of integration, and performing the integration. Suitable for students in Grades 11-12.

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