the region in the first quadrant that is bounded above by the curve y = 1/ (Sqrtx) on the left by the line x = 1/4 and below by the line y = 1 is revolved about the y axis to generate a solid. find the volume of the solid 

We are tasked with finding the volume of the solid formed when the region bounded by the curve \( y = \frac{1}{\sqrt{x}} \), the line \( x = \frac{1}{4} \), and the line \( y = 1 \), in the first quadrant, is revolved about the y-axis.

### Step 1: Rearranging the curve equation
The curve is given as \( y = \frac{1}{\sqrt{x}} \). To express \( x \) in terms of \( y \), solve for \( x \):

\[
y = \frac{1}{\sqrt{x}} \quad \Rightarrow \quad \sqrt{x} = \frac{1}{y} \quad \Rightarrow \quad x = \frac{1}{y^2}.
\]

Thus, the curve can also be expressed as \( x = \frac{1}{y^2} \).

### Step 2: Setting up the limits
We are rotating the region bounded by the curve, the line \( x = \frac{1}{4} \), and the line \( y = 1 \) around the y-axis. The volume of the solid can be computed using the method of cylindrical shells or the disk/washer method. Here, we will use the disk/washer method.

- The limits for \( y \) are from \( y = 1 \) to the value of \( y \) when \( x = \frac{1}{4} \).
  
When \( x = \frac{1}{4} \), substitute into \( y = \frac{1}{\sqrt{x}} \):

\[
y = \frac{1}{\sqrt{\frac{1}{4}}} = 2.
\]

Thus, the limits for \( y \) are from \( y = 1 \) to \( y = 2 \).

### Step 3: Volume formula using the disk method
The volume \( V \) of the solid of revolution is given by:

\[
V = \pi \int_{y_1}^{y_2} [R(y)]^2 \, dy,
\]

where \( R(y) \) is the distance from the y-axis to the boundary of the region, which in this case is \( R(y) = \frac{1}{y^2} \).

Thus, the volume integral becomes:

\[
V = \pi \int_{1}^{2} \left( \frac{1}{y^2} \right)^2 \, dy.
\]

### Step 4: Simplifying and solving the integral
Simplifying \( \left( \frac{1}{y^2} \right)^2 = \frac{1}{y^4} \), the volume expression becomes:

\[
V = \pi \int_{1}^{2} \frac{1}{y^4} \, dy.
\]

Now, integrate \( \frac{1}{y^4} \):

\[
\int \frac{1}{y^4} \, dy = -\frac{1}{3y^3}.
\]

Now, evaluate this from \( y = 1 \) to \( y = 2 \):

\[
V = \pi \left[ -\frac{1}{3y^3} \right]_1^2 = \pi \left( -\frac{1}{3(2)^3} + \frac{1}{3(1)^3} \right)
= \pi \left( -\frac{1}{24} + \frac{1}{3} \right)
= \pi \left( \frac{1}{3} - \frac{1}{24} \right)
= \pi \cdot \frac{8}{24}
= \frac{2\pi}{3}.
\]

### Final Answer:
The volume of the solid is \( \frac{2\pi}{3} \) cubic units.

---

Would you like a more detailed breakdown, or do you have any further questions?

Here are 5 related questions for you:
1. How does the disk method differ from the shell method for volume calculation?
2. What would change if the region were revolved about the x-axis instead?
3. How do we determine the limits of integration for a solid of revolution?
4. What would happen if the region were revolved around a different vertical line, like \( x = 1 \)?
5. How would the integral change if the curve were \( y = \frac{1}{x} \)?

**Tip**: When setting up volume integrals, always carefully identify the boundaries and the axis of revolution to choose the right method (disk/washer or shell).

The region in the first quadrant that is bounded above by the curve y = 1/√x, on the left by the line x = 1/4, and below by the line y = 1 is revolved about the y-axis to generate a solid. Find the volume of the solid.

Calculate the volume of a solid formed by revolving the region bounded by y = 1/√x, x = 1/4, and y = 1 about the y-axis. This problem applies calculus concepts like the disk method and volume by integration, suitable for undergraduate calculus students.

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