The integral gives the area of the region in the​ xy-plane. Sketch the​ region, label each bounding curve with its​ equation, and give the coordinates of the points where the curves intersect. Then find the area of the region.

Integral from nothing to nothing Subscript 0 Superscript 24∫240Integral from nothing to nothing Subscript y squared divided by 6 Superscript 4 y Baseline dx font size decreased by 1 font size decreased by 1 dy

To solve this problem, we need to clarify and interpret the integral given. From your description, it seems like you're dealing with a double integral, which can be written as follows:

\[
\int_{0}^{4} \int_{\frac{y^2}{6}}^{4y} dx \, dy
\]

### Step 1: Understand the integral and the region

- **Outer integral:** \( \int_{0}^{4} dy \), meaning \( y \) goes from 0 to 4.
- **Inner integral:** \( \int_{\frac{y^2}{6}}^{4y} dx \), meaning for each \( y \), \( x \) goes from \( \frac{y^2}{6} \) to \( 4y \).

### Step 2: Identify the bounding curves

We are given two bounding curves for \( x \):
- \( x = \frac{y^2}{6} \)
- \( x = 4y \)

### Step 3: Find points of intersection

To find the points where these curves intersect, we set the equations equal to each other:
\[
\frac{y^2}{6} = 4y
\]
Multiplying through by 6 to clear the denominator:
\[
y^2 = 24y
\]
Rearrange the equation:
\[
y^2 - 24y = 0
\]
Factor the quadratic:
\[
y(y - 24) = 0
\]
Thus, \( y = 0 \) or \( y = 24 \).

- When \( y = 0 \), substitute into either equation for \( x \) (since both curves pass through this point):
  - \( x = \frac{0^2}{6} = 0 \)
  - \( x = 4(0) = 0 \)

So, one point of intersection is \( (0, 0) \).

- When \( y = 24 \), substitute into either equation for \( x \):
  - \( x = \frac{24^2}{6} = 96 \)
  - \( x = 4(24) = 96 \)

So, the second point of intersection is \( (96, 24) \).

### Step 4: Sketch the region

The region is bounded by the curves \( x = \frac{y^2}{6} \) and \( x = 4y \), from \( y = 0 \) to \( y = 24 \). The area of the region lies between these two curves, and the limits of integration cover the entire region in the first quadrant.

### Step 5: Calculate the area

The double integral we are evaluating represents the area between these curves. We perform the integration as follows:

#### Step 5a: Inner integral
\[
\int_{\frac{y^2}{6}}^{4y} dx = \left[ x \right]_{\frac{y^2}{6}}^{4y} = 4y - \frac{y^2}{6}
\]

#### Step 5b: Outer integral
Now, integrate this expression with respect to \( y \) from 0 to 24:
\[
\int_{0}^{24} \left( 4y - \frac{y^2}{6} \right) dy
\]
Split the integral:
\[
\int_{0}^{24} 4y \, dy - \int_{0}^{24} \frac{y^2}{6} \, dy
\]

#### Step 5c: Solve each part
- For the first integral:
  \[
  \int_{0}^{24} 4y \, dy = 4 \left[ \frac{y^2}{2} \right]_{0}^{24} = 4 \times \frac{24^2}{2} = 4 \times 288 = 1152
  \]
- For the second integral:
  \[
  \int_{0}^{24} \frac{y^2}{6} \, dy = \frac{1}{6} \left[ \frac{y^3}{3} \right]_{0}^{24} = \frac{1}{6} \times \frac{24^3}{3} = \frac{1}{6} \times \frac{13824}{3} = 768
  \]

#### Step 5d: Subtract the results
\[
1152 - 768 = 384
\]

### Conclusion

The area of the region is **384 square units**.

Would you like further details on any part of the solution? Here are five follow-up questions:

1. What is a double integral and when is it used?
2. How do you interpret the bounds of a double integral?
3. How can you solve double integrals geometrically?
4. What are the applications of double integrals in physics and engineering?
5. How can you verify the area using another method, such as geometry?

**Tip:** When setting up double integrals, always check that the limits of integration match the region you're describing.

The integral gives the area of the region in the xy-plane. Sketch the region, label each bounding curve with its equation, and give the coordinates of the points where the curves intersect. Then find the area of the region. Integral from 0 to 4 ∫ (y²/6) to 4y dx dy

Learn how to calculate the area between the curves x = y²/6 and x = 4y using double integrals. This problem covers key calculus concepts such as curve intersection, bounding regions, and double integration techniques. Suitable for college-level calculus students.

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