Let R be the region bounded by the graphs y =
√
x − 1 and y = (x − 1)2
.
(a) Set up a integral for the area of the region R in terms of y

To set up an integral for the area of the region \( R \) bounded by the curves \( y = \sqrt{x - 1} \) and \( y = (x - 1)^2 \), we will first express the area in terms of \( y \).

### Step 1: Express \( x \) in terms of \( y \) for both functions.

- For \( y = \sqrt{x - 1} \), solve for \( x \):
  \[
  y = \sqrt{x - 1} \implies y^2 = x - 1 \implies x = y^2 + 1
  \]

- For \( y = (x - 1)^2 \), solve for \( x \):
  \[
  y = (x - 1)^2 \implies \sqrt{y} = x - 1 \implies x = \sqrt{y} + 1
  \]

### Step 2: Find the limits of integration in terms of \( y \).

The region \( R \) is bounded where the two curves intersect. Set the two expressions for \( y \) equal to each other to find the points of intersection:
\[
\sqrt{x - 1} = (x - 1)^2
\]
Let \( u = x - 1 \), then:
\[
\sqrt{u} = u^2
\]
Square both sides:
\[
u = u^4
\]
Rearrange:
\[
u^4 - u = 0 \implies u(u^3 - 1) = 0
\]
Thus, \( u = 0 \) or \( u = 1 \). This gives:
\[
x - 1 = 0 \implies x = 1 \quad \text{and} \quad x - 1 = 1 \implies x = 2
\]
So, the points of intersection are \( x = 1 \) and \( x = 2 \). Since the curves are written in terms of \( x - 1 \), in terms of \( y \), the values range from \( y = 0 \) to \( y = 1 \).

### Step 3: Set up the integral for the area.

The area between the curves can now be computed by integrating the difference between the \( x \)-values of the two functions, expressed in terms of \( y \):
\[
\text{Area} = \int_{0}^{1} \left[ (\sqrt{y} + 1) - (y^2 + 1) \right] \, dy
\]
Simplify the integrand:
\[
\text{Area} = \int_{0}^{1} \left( \sqrt{y} - y^2 \right) \, dy
\]

Thus, the integral for the area of the region \( R \) in terms of \( y \) is:
\[
\boxed{\int_{0}^{1} \left( \sqrt{y} - y^2 \right) \, dy}
\]

Would you like me to solve the integral, provide more details, or clarify any part of the setup?

### 5 Related Questions:
1. How do you solve this integral for the exact area?
2. How would the problem change if you set up the integral with respect to \( x \) instead of \( y \)?
3. Can you find the centroid of the region \( R \)?
4. What if the region was rotated around the \( y \)-axis? How would you set up the volume integral?
5. How can you compute the perimeter of the region?

### Tip:
When switching between variables (from \( x \) to \( y \) or vice versa), always carefully solve for the limits of integration and check the intersection points.

Let R be the region bounded by the graphs y = √(x - 1) and y = (x - 1)^2. (a) Set up an integral for the area of the region R in terms of y.

Learn how to set up a definite integral to find the area of the region bounded by the curves y = √(x - 1) and y = (x - 1)^2. This calculus problem involves key concepts such as definite integrals, finding intersections, and integrating with respect to y. Suitable for undergraduate calculus students.

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