Question Statement

Find the area of the region enclosed by $x=y^2$ and $y=x-2$ integrating with respect to $y$.

---

Background and Explanation

When finding areas between curves, integrating with respect to $y$ is advantageous when the region is bounded by functions of the form $x=f(y)$. This approach uses horizontal strips where the width is the difference between the rightmost and leftmost $x$-values at each $y$.

---

Solution

First, express both curves as functions of $x$ in terms of $y$:

• The parabola: $x=y^2$
• The line: From $y=x-2$, we get $x=y+2$

Finding the limits of integration:

To find where the curves intersect, set the $x$-values equal: $y^2=y+2$

Rearranging into standard quadratic form: $y^2-y-2=0$

Factoring the quadratic: $(y-2)(y+1)=0$

Therefore, the curves intersect at $y=2$ and $y=-1$. These are our limits of integration.

Setting up the integral:

For $y\in [-1,2]$, test a point (say $y=0$) to determine which curve is on the right:

• Line: $x=0+2=2$
• Parabola: $x=0^2=0$

Thus, $x=y+2$ is the right curve and $x=y^2$ is the left curve. The area is:

$\text{Area}={\int }_{-1}^2\left[(y+2)-y^2\right]dy={\int }_{-1}^2(y+2-y^2)dy$

Evaluating the integral:

Split the integral into three separate terms for easier computation: $\text{Area}={\int }_{-1}^{2}ydy+2{\int }_{-1}^{2}1dy-{\int }_{-1}^2{y}^{2}dy$

Compute each definite integral using the Fundamental Theorem of Calculus:

1. For ${\int }_{-1}^{2}ydy$: ${\left[\frac{y^2}{2}\right]}_{-1}^2=\frac{1}{2}\left[(2{)}^2-(-1{)}^2\right]=\frac{1}{2}(4-1)=\frac{3}{2}$

2. For $2{\int }_{-1}^{2}1dy$: $2{\left[y\right]}_{-1}^2=2\left[2-(-1)\right]=2(3)=6$

3. For ${\int }_{-1}^2{y}^{2}dy$: ${\left[\frac{y^3}{3}\right]}_{-1}^2=\frac{1}{3}\left[(2{)}^3-(-1{)}^3\right]=\frac{1}{3}(8-(-1))=\frac{1}{3}(9)=3$

Combining the results:

$\text{Area}=\frac{3}{2}+6-3=\frac{3}{2}+3=\frac{3+6}{2}=\frac{9}{2}$

Therefore, the area of the enclosed region is $\boxed{\frac{9}{2}}$ square units.

---

Key Formulas or Methods Used

• Area between curves (horizontal strips): $A={\int }_c^d\left[x_{\text{right}}(y)-x_{\text{left}}(y)\right]dy$
• Finding intersection points: Solve simultaneous equations by setting $x$-values equal
• Power rule for integration: $\int y^{n}dy=\frac{y^{n+1}}{n+1}+C$ (for $n\ne -1$)
• Fundamental Theorem of Calculus: ${\int }_a^{b}f(y)dy=F(b)-F(a)$ where $F$ is the antiderivative of $f$

---

Summary of Steps

1. Rewrite both equations as $x$ in terms of $y$: $x=y^2$ and $x=y+2$
2. Find intersection points by solving $y^2=y+2$, yielding $y=-1$ and $y=2$
3. Identify the right curve ($x=y+2$) and left curve ($x=y^2$) by testing a point between the intersection values
4. Set up the area integral: ${\int }_{-1}^2\left[(y+2)-y^2\right]dy$
5. Expand and integrate term by term using the power rule
6. Evaluate the antiderivatives at the bounds $y=2$ and $y=-1$
7. Combine the numerical results to obtain the final answer: $\frac{9}{2}$ square units