find the area of the region: 2x + y^2 = 8,  x = y

To find the area of the region enclosed by the curves \( 2x + y^2 = 8 \) and \( x = y \), we will proceed step by step:

### Step 1: Solve for \( x \) in terms of \( y \)
Start with the equation of the first curve:
\[
2x + y^2 = 8
\]
Solve for \( x \):
\[
2x = 8 - y^2 \quad \Rightarrow \quad x = \frac{8 - y^2}{2}
\]

Now, we have the equations of the two curves:
1. \( x = \frac{8 - y^2}{2} \)
2. \( x = y \)

### Step 2: Find the points of intersection
To find the points of intersection, set the two expressions for \( x \) equal to each other:
\[
\frac{8 - y^2}{2} = y
\]
Multiply both sides by 2 to eliminate the denominator:
\[
8 - y^2 = 2y
\]
Rearrange the equation:
\[
y^2 + 2y - 8 = 0
\]
Solve this quadratic equation using the quadratic formula:
\[
y = \frac{-2 \pm \sqrt{(2)^2 - 4(1)(-8)}}{2(1)} = \frac{-2 \pm \sqrt{4 + 32}}{2} = \frac{-2 \pm \sqrt{36}}{2} = \frac{-2 \pm 6}{2}
\]
Thus, the solutions are:
\[
y = \frac{-2 + 6}{2} = 2 \quad \text{or} \quad y = \frac{-2 - 6}{2} = -4
\]

So the points of intersection are at \( y = 2 \) and \( y = -4 \).

### Step 3: Set up the integral for the area
The area between the two curves is given by the integral of the difference between the right-hand curve \( x = \frac{8 - y^2}{2} \) and the left-hand curve \( x = y \) over the interval from \( y = -4 \) to \( y = 2 \).

The area \( A \) is:
\[
A = \int_{-4}^{2} \left( \frac{8 - y^2}{2} - y \right) \, dy
\]

### Step 4: Simplify the integrand
Simplify the expression inside the integral:
\[
\frac{8 - y^2}{2} - y = \frac{8}{2} - \frac{y^2}{2} - y = 4 - \frac{y^2}{2} - y
\]

Now, the integral becomes:
\[
A = \int_{-4}^{2} \left( 4 - \frac{y^2}{2} - y \right) \, dy
\]

### Step 5: Compute the integral
We can now compute the integral term by term:
\[
A = \int_{-4}^{2} 4 \, dy - \int_{-4}^{2} \frac{y^2}{2} \, dy - \int_{-4}^{2} y \, dy
\]

- The first term:
\[
\int_{-4}^{2} 4 \, dy = 4[y]_{-4}^{2} = 4(2 - (-4)) = 4 \times 6 = 24
\]

- The second term:
\[
\int_{-4}^{2} \frac{y^2}{2} \, dy = \frac{1}{2} \int_{-4}^{2} y^2 \, dy = \frac{1}{2} \left[ \frac{y^3}{3} \right]_{-4}^{2} = \frac{1}{2} \left( \frac{(2)^3}{3} - \frac{(-4)^3}{3} \right) = \frac{1}{2} \left( \frac{8}{3} - \frac{-64}{3} \right) = \frac{1}{2} \times \frac{72}{3} = 12
\]

- The third term:
\[
\int_{-4}^{2} y \, dy = \left[ \frac{y^2}{2} \right]_{-4}^{2} = \frac{(2)^2}{2} - \frac{(-4)^2}{2} = \frac{4}{2} - \frac{16}{2} = 2 - 8 = -6
\]

### Step 6: Add the results
Now, sum the results of the integrals:
\[
A = 24 - 12 - (-6) = 24 - 12 + 6 = 18
\]

Thus, the area of the region is \( 18 \) square units.

### Would you like a more detailed breakdown of any step?

---

Here are five related questions to explore further:
1. How would the area change if the curve \( x = y^2 \) was involved instead of \( x = y \)?
2. Can you find the area using horizontal slices instead of vertical ones?
3. What is the geometric shape of the region enclosed by these curves?
4. How would the solution differ if the bounds of integration changed?
5. What if the equation of the curve was \( 2x + y^3 = 8 \) instead of \( y^2 \)?

**Tip**: Always simplify the integrand as much as possible before calculating the integral to make the process easier!

Find the area of the region: 2x + y^2 = 8, x = y

This detailed solution explains how to find the area of the region enclosed by the curves 2x + y^2 = 8 and x = y using integral calculus. Learn key concepts such as setting up integrals, solving quadratic equations, and applying the Fundamental Theorem of Calculus. Suitable for Grades 11-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation