Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis. y=2sqrt(x), x=0 
 about y=3

To find the volume of the solid obtained by rotating the region bounded by \( y = 2\sqrt{x} \) and \( x = 0 \) about the line \( y = 3 \), we use the **method of cylindrical shells** or **washer method**, depending on the axis of rotation.

### Step 1: Identify the region
The curve given is \( y = 2\sqrt{x} \), which implies that \( x = \frac{y^2}{4} \). The region is bounded by \( x = 0 \) (the y-axis), which occurs at \( y = 0 \), and extends vertically to some value where \( x = 0 \).

Since we're rotating about \( y = 3 \) (a horizontal line), we will use the **washer method** because the axis is parallel to the \( x \)-axis.

### Step 2: Washer method formula

The volume \( V \) is given by:

\[
V = \pi \int_{y_1}^{y_2} \left[ (R_{\text{outer}}(y))^2 - (R_{\text{inner}}(y))^2 \right] dy
\]

Where:
- \( R_{\text{outer}}(y) \) is the distance from the axis \( y = 3 \) to the outer curve (in this case, the x-axis),
- \( R_{\text{inner}}(y) \) is the distance from the axis \( y = 3 \) to the curve \( y = 2\sqrt{x} \).

### Step 3: Set up the radii

- The **outer radius** is the distance from the axis \( y = 3 \) to the line \( y = 0 \) (y-axis):
  \[
  R_{\text{outer}} = 3 - 0 = 3
  \]

- The **inner radius** is the distance from the axis \( y = 3 \) to the curve \( y = 2\sqrt{x} \), which can be written in terms of \( y \):
  \[
  R_{\text{inner}} = 3 - y
  \]

### Step 4: Limits of integration

The curve \( y = 2\sqrt{x} \) intersects the x-axis at \( y = 0 \). The maximum value of \( y \) occurs at the point where the curve reaches \( x = 0 \), which is at \( y = 2 \).

So, the limits of integration are from \( y_1 = 0 \) to \( y_2 = 2 \).

### Step 5: Set up the integral

The volume integral is:

\[
V = \pi \int_0^2 \left[ 3^2 - (3 - y)^2 \right] dy
\]

Expanding the terms inside the integral:

\[
V = \pi \int_0^2 \left[ 9 - (9 - 6y + y^2) \right] dy
\]

Simplifying:

\[
V = \pi \int_0^2 \left[ 9 - 9 + 6y - y^2 \right] dy
\]

\[
V = \pi \int_0^2 \left[ 6y - y^2 \right] dy
\]

### Step 6: Compute the integral

First, integrate \( 6y \) and \( y^2 \):

\[
\int_0^2 6y \, dy = 3y^2 \Big|_0^2 = 3(2^2) - 3(0^2) = 12
\]
\[
\int_0^2 y^2 \, dy = \frac{y^3}{3} \Big|_0^2 = \frac{2^3}{3} - \frac{0^3}{3} = \frac{8}{3}
\]

Now subtract:

\[
V = \pi \left( 12 - \frac{8}{3} \right) = \pi \left( \frac{36}{3} - \frac{8}{3} \right) = \pi \cdot \frac{28}{3}
\]

\[
V = \frac{28\pi}{3}
\]

### Final Answer:
The volume of the solid is \( \frac{28\pi}{3} \) cubic units.

---

Would you like more details on any part of the process?

Here are 5 related questions:
1. What changes if the region is rotated around the x-axis instead of \( y = 3 \)?
2. How do we handle the washer method for regions bounded by two curves?
3. Can the volume be calculated using the shell method for this problem?
4. How does the axis of rotation affect the choice of method for volume calculation?
5. What happens if the curve extends beyond \( y = 2 \)?

**Tip:** When solving rotation problems, always sketch the region and axis of rotation to visualize the setup better.

Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis. y=2sqrt(x), x=0 about y=3

Learn how to calculate the volume of a solid formed by rotating the curve y=2√x about y=3. This problem uses the washer method, with detailed steps on setting up the integral, finding limits of integration, and solving the volume. Suitable for AP Calculus or college-level students.

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