# How to Find the Area Enclosed by Curves Using Any Axes

Finding the area between curves is a fundamental concept in integral calculus that involves computing the region enclosed by two or more functions over a specific interval. This is achieved by integrating the difference between the functions. The method depends on how the functions are defined: if they are functions of \( x \), use vertical slices and integrate with respect to \( x \); if they are functions of \( y \), use horizontal slices and integrate with respect to \( y \). Identifying which function is uppermost (or rightmost) in the interval is crucial for setting up the correct integral. This technique has wide applications in physics, engineering, and economics.

To find the area between curves, choose the axis of integration carefully. Use vertical slices \(dx\) for functions defined as \( y = f(x) \) and \(y = g(x) \), integrating \([f(x) − g(x)]\) from \(x = a\) to \(x = b\). Use horizontal slices \(dy\) for \(x = f(y)\) and \(x = g(y)\), integrating \([f(y) − g(y)]\) from \(y = c\) to \(y = d\). When curves intersect multiple times, identify which curve is on top in each segment, set up separate integrals, and sum the areas.

For more advanced cases, when dealing with parametric or polar equations, the area between curves requires specialized integration techniques. For parametric curves \( (x(t), y(t)) \), the area is calculated using \( \int_{t_1}^{t_2} [x(t) y'(t) – y(t) x'(t)] \, dt \). In polar coordinates, the area between curves is found using \( \frac{1}{2} \int_{\theta_1}^{\theta_2} [r_2^2 – r_1^2] \, d\theta \).

Additionally, when dealing with implicit functions or regions bounded by complex curves, double integrals are used to compute the area. By setting up a double integral over the region \( R \), the area is \( \iint_R dA \). Green’s Theorem can also be applied to find areas by converting a line integral around a closed curve into a double integral over the region it encloses, which is especially useful in vector calculus.

**Example: Finding the Area Between Curves Using Vertical and Horizontal Slices**

**Problem:**

Find the area of the region enclosed by the curves \( y = x^2 \) and \( y = 2x \).

**Solution:**

**Step 1: Sketch the Curves**

First, sketch the graphs of the functions to understand the region we’re dealing with.

- \( y = x^2 \) is a parabola opening upwards.
- \( y = 2x \) is a straight line.

**Step 2: Find Points of Intersection**

Set the equations equal to find the points where the curves intersect.

\( x^2 – 2x =0 \)

So, \( x = 0 \) or \( x = 2 \).

**Step 3: Determine Which Curve Is on Top**

For \( x \) between \( 0 \) and \(2\):

- Choose \( x = 1 \):
- \( y = x^2 = 1^2 = 1 \)
- \( y = 2x = 2 \times 1 = 2 \)

Since \( y = 2x \) yields a higher value, \( y = 2x \) is the upper function, and \( y = x^2 \) is the lower function in this interval.

**Step 4: Set Up the Integral Using Vertical Slices (dx)**

The area \( A \) between the curves from \( x = 0 \) to \( x = 2 \) is:

\( [A = \int_{0}^{2} [\text{Upper function} – \text{Lower function}] \, dx = \int_{0}^{2} [2x – x^2] \, dx]\)

**Step 5: Evaluate the Integral**

Compute the integral:

\( [\begin{align} A &= \int_{0}^{2} (2x – x^2) \, dx \ &= \left[ x^2 – \frac{1}{3} x^3 \right]_0^2 \ &= \left( 2^2 – \frac{1}{3} \times 2^3 \right) – \left( 0^2 – \frac{1}{3} \times 0^3 \right) \ &= \left( 4 – \frac{8}{3} \right) – 0 \ &= \left( \frac{12}{3} – \frac{8}{3} \right) \ &= \frac{4}{3} \end{align}] \)

**Answer:**

The area of the region enclosed by \( y = x^2 \) and \( y = 2x \) is \( \frac{4}{3} \) square units.

**Alternative Scenario with Horizontal Slices \(dy\):**

Suppose we have functions defined in terms of \( y \):

**Problem:**

Find the area of the region enclosed by \( x = y^2 \) and \( x = y + 2 \).

**Solution:**

**Step 1: Sketch the Curves**

- \( x = y^2 \) is a parabola opening to the right.
- \( x = y + 2 \) is a straight line.

**Step 2: Find Points of Intersection**

Set the equations equal:

\( [\begin{align} y^2 &= y + 2 \ y^2 – y – 2 &= 0 \ (y – 2)(y + 1) &= 0 \end{align}]\)

So, \( y = -1 \) or \( y = 2 \).

**Step 3: Determine Which Curve Is on the Right**

For \( y \) between \( -1 \) and \( 2 \):

- Choose \( y = 0 \):
- \( x = y^2 = 0^2 = 0 \)
- \( x = y + 2 = 0 + 2 = 2 \)

Since \( x = y + 2 \) yields a higher value, it’s the rightmost function.

**Step 4: Set Up the Integral Using Horizontal Slices \(dy\)**

The area \( A \) is:

\( [A = \int_{-1}^{2} [\text{Rightmost function} – \text{Leftmost function}] \, dy = \int_{-1}^{2} [(y + 2) – y^2] ,dy] \)

**Step 5: Evaluate the Integral**

Compute the integral:

\( [\begin{align} A &= \int_{-1}^{2} (y + 2 – y^2) \, dy \ &= \int_{-1}^{2} (-y^2 + y + 2) \, dy \ &= \left[ -\frac{1}{3} y^3 + \frac{1}{2} y^2 + 2y \right]_{-1}^{2} \ \end{align}]\)

Evaluate at \( y = 2 \):

\( [\left( -\frac{1}{3} \times 8 + \frac{1}{2} \times 4 + 4 \right) = \left( -\frac{8}{3} + 2 + 4 \right) = \frac{10}{3}] \)

Evaluate at \( y = -1 \):

\( [\left( -\frac{1}{3} \times (-1) + \frac{1}{2} \times 1 – 2 \right) = \left( \frac{1}{3} + \frac{1}{2} – 2 \right) = -\frac{7}{6}] \)

Subtract:

\( [A = \frac{10}{3} – \left( -\frac{7}{6} \right) = \frac{10}{3} + \frac{7}{6} = \frac{27}{6} = \frac{9}{2}] \)

**Answer:**

The area of the region enclosed by \( x = y^2 \) and \( x = y + 2 \) is \( \frac{9}{2} \) square units.

**Handling Multiple Intersections:**

**Problem:**

Find the area between the curves \( y = x^2 \) and \( y = (x – 2)^2 \).

**Solution:**

**Step 1: Find Points of Intersection**

Set \( x^2 = (x – 2)^2 \):

\( [\begin{align} x^2 &= x^2 – 4x + 4 \ 0 &= -4x + 4 \ x &= 1 \end{align}] \)

They intersect at \( x = 1 \).

**Step 2: Determine Regions of Integration**

Since the curves intersect at \( x = 1 \) and are symmetric, we’ll consider the interval from \( x = 0 \) to \( x = 2 \), splitting at \( x = 1 \).

**For \( x = 0 \) to \( x = 1 \):**

- \( y = (x – 2)^2 \) is above \( y = x^2 \).

**For \( x = 1 \) to \( x = 2 \):**

- \( y = x^2 \) is above \( y = (x – 2)^2 \).

**Step 3: Set Up the Integrals**

**First Integral \(( x = 0 ) to ( x = 1 )\):**

\( [A_1 = \int_{0}^{1} [(x – 2)^2 – x^2] \, dx]\)

Simplify:

\([(x – 2)^2 – x^2 = x^2 – 4x + 4 – x^2 = -4x + 4]\)

So,

\([A_1 = \int_{0}^{1} (-4x + 4) \, dx]\)

**Second Integral (( x = 1 ) to ( x = 2 )):**

\([A_2 = \int_{1}^{2} [x^2 – (x – 2)^2] \, dx]\)

Simplify:

\([x^2 – (x – 2)^2 = x^2 – (x^2 – 4x + 4) = 4x – 4]\)

So,

\([A_2 = \int_{1}^{2} (4x – 4) \, dx]\)

**Step 4: Evaluate the Integrals**

**First Integral:**

\([\begin{align} A_1 &= \int_{0}^{1} (-4x + 4) \, dx \ &= \left[ -2x^2 + 4x \right]_0^1 \ &= \left( -2 \times 1^2 + 4 \times 1 \right) – \left( -2 \times 0^2 + 4 \times 0 \right) \ &= (-2 + 4) – 0 = 2 \end{align}]\)

**Second Integral:**

\([\begin{align} A_2 &= \int_{1}^{2} (4x – 4) \, dx \ &= \left[ 2x^2 – 4x \right]_1^2 \ &= \left( 2 \times 2^2 – 4 \times 2 \right) – \left( 2 \times 1^2 – 4 \times 1 \right) \ &= (8 – 8) – (2 – 4) = 0 – (-2) = 2 \end{align}] \)

**Step 5: Sum the Areas**

\([A = A_1 + A_2 = 2 + 2 = 4]\)

**Answer:**

The total area between \( y = x^2 \) and \( y = (x – 2)^2 \) from \( x = 0 \) to \( x = 2 \) is \( 4 \) square units.

**Conclusion:**

In each example, we carefully chose the axis of integration based on how the functions were defined and identified which curve was on top (or to the right) in each interval. When the curves intersected multiple times, we split the integrals accordingly and summed the areas to find the total area between the curves.

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