Find the area of the surface generated by revolving x equals 2 StartRoot 1 minus y EndRoot​, 0 less than or equals y less than or equals three fourths
 about the​ y-axis.

The problem asks us to find the surface area of a solid generated by revolving the curve \( x = 2 \sqrt{1 - y} \) for \( 0 \leq y \leq \frac{3}{4} \) around the y-axis. We can use the formula for the surface area of a solid of revolution about the y-axis:

\[
A = 2 \pi \int_{a}^{b} x \sqrt{1 + \left( \frac{dx}{dy} \right)^2} \, dy
\]

### Step 1: Expression for \( x \)
The curve is given by \( x = 2 \sqrt{1 - y} \).

### Step 2: Differentiate \( x \) with respect to \( y \)
We need to find \( \frac{dx}{dy} \):

\[
\frac{dx}{dy} = 2 \cdot \frac{1}{2} \cdot (1 - y)^{-\frac{1}{2}} \cdot (-1) = -\frac{1}{\sqrt{1 - y}}
\]

### Step 3: Set up the formula for surface area
Now, substitute \( x = 2 \sqrt{1 - y} \) and \( \frac{dx}{dy} = -\frac{1}{\sqrt{1 - y}} \) into the surface area formula:

\[
A = 2 \pi \int_0^{\frac{3}{4}} 2 \sqrt{1 - y} \cdot \sqrt{1 + \left( -\frac{1}{\sqrt{1 - y}} \right)^2} \, dy
\]

Simplify the expression inside the square root:

\[
\sqrt{1 + \left( -\frac{1}{\sqrt{1 - y}} \right)^2} = \sqrt{1 + \frac{1}{1 - y}} = \sqrt{\frac{1}{1 - y}}
\]

So the integral becomes:

\[
A = 2 \pi \int_0^{\frac{3}{4}} 2 \sqrt{1 - y} \cdot \frac{1}{\sqrt{1 - y}} \, dy
\]

The \( \sqrt{1 - y} \) terms cancel out, leaving:

\[
A = 4 \pi \int_0^{\frac{3}{4}} 1 \, dy
\]

### Step 4: Evaluate the integral
The integral is straightforward:

\[
A = 4 \pi \left[ y \right]_0^{\frac{3}{4}} = 4 \pi \left( \frac{3}{4} - 0 \right) = 3 \pi
\]

### Final Answer:
The surface area is \( 3 \pi \) square units.

Would you like more details or clarifications?

Here are some related questions:
1. How do you compute the surface area if revolving around the x-axis?
2. What would happen if the limits of integration changed?
3. How do we adapt the formula for a different curve?
4. What role does the derivative \( \frac{dx}{dy} \) play in surface area computation?
5. How does this problem differ if \( y \) is revolved around other axes?

**Tip:** Always check that the limits of integration match the range of the variable you're revolving around.

Find the area of the surface generated by revolving x equals 2 sqrt(1 - y), 0 less than or equals y less than or equals three fourths about the y-axis.

Learn how to calculate the surface area generated by revolving the curve x = 2√(1 - y) for 0 ≤ y ≤ 3/4 around the y-axis. This solution covers calculus concepts such as the surface area of revolution and integration, suitable for students in Grades 11-12.

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