How to Find the Area Enclosed by Curves Using Any Axes

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.

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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.

Frequently Asked Questions

How do you add and subtract mixed fractions?

Adding and subtracting mixed fractions involves a few clear steps. First, ensure that the fractions have a common denominator, much like you would when finding the area between curves by ensuring the equations are compared over the same interval. Convert each mixed fraction to an improper fraction, where the numerator is greater than the denominator. Then, add or subtract the numerators as required. Finally, simplify the resulting fraction if possible. For more detailed steps and examples, you might find our page on fraction fundamentals helpful.

How do you add and subtract decimals?

To add and subtract decimals, start by aligning the numbers by their decimal points, ensuring that each digit is in the correct place value column — units, tenths, hundredths, etc. When adding, simply combine the numbers in each column starting from the rightmost, carrying over as needed. For subtraction, subtract each column from right to left, borrowing if necessary. This method ensures accuracy, much like how we find the area between curves in calculus by integrating with respect to the correct variable, ensuring the calculations respect the placement and relationship between the functions involved. For additional practice with decimals, you can explore resources on EffortlessMath.com.

How do you find the area of a triangle?

To find the area of a triangle, you’ll use the formula \( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \). This involves measuring the length of the base of the triangle and the height, which is the perpendicular distance from the base to the opposite vertex. These measurements are then substituted into the formula. This mathematical concept is a simpler, specific case of finding the area between curves, where the curves are straight lines forming the sides of the triangle. If you’re interested in broader applications of area calculations, such as finding the area between more complex curves, exploring topics in integral calculus might be helpful.