AP Calculus BC 8.5 Finding the Area Between Curves Expressed as Functions of y Study Notes - New Syllabus
AP Calculus BC 8.5 Finding the Area Between Curves Expressed as Functions of y Study Notes- New syllabus
AP Calculus BC 8.5 Finding the Area Between Curves Expressed as Functions of y Study Notes – AP Calculus BC- per latest AP Calculus BC Syllabus.
LEARNING OBJECTIVE
- Definite integrals allow us to solve problems involving the accumulation of change over an interval.
Key Concepts:
- Finding the Area Between Curves Expressed as Functions of y
Finding the Area Between Curves Expressed as Functions of y
Area Under the Curve and the y-axis
When the region is bounded by a curve \( x = g(y) \) and the y-axis, we compute the area using integration with respect to \( y \).
- If \( g(y) \geq 0 \) on \( [c, d] \), the definite integral \( \displaystyle \int_{c}^{d} g(y) \, dy \) gives the actual area between the curve and the y-axis.
- If \( g(y) \leq 0 \), the definite integral will be negative; the actual area is the absolute value of the result.
- If \( g(y) \) changes sign in \( [c, d] \), split the interval at \( g(y) = 0 \) and integrate absolute values over each part.
Total Area Formula:
\( A_{\text{total}} = \displaystyle \int_{c}^{d} |g(y)| \, dy \)
Procedure:
- Express the curve in the form \( x = g(y) \).
- Find where \( g(y) = 0 \) to identify points where the curve touches the y-axis.
- Split the range of \( y \) into intervals where \( g(y) \) has a consistent sign.
- Integrate \( |g(y)| \) over each interval and sum to get the total area.
Example:
Find the total area enclosed between \( x = y^2 – 4 \) and the y-axis for \( -2 \leq y \leq 2 \).
▶️ Answer/Explanation
Here \( g(y) = y^2 – 4 \).
On \( [-2, 2] \), \( y^2 – 4 \leq 0 \) because \( y^2 \leq 4 \).
Total area:
\( A_{\text{total}} = \displaystyle \int_{-2}^{2} |y^2 – 4| \, dy \)
Since \( y^2 – 4 \leq 0 \), \(|y^2 – 4| = 4 – y^2\).
Evaluate: \( A_{\text{total}} = \displaystyle \int_{-2}^{2} (4 – y^2) \, dy \)
\( = \left[ 4y – \dfrac{y^3}{3} \right]_{-2}^{2} = \left(8 – \dfrac{8}{3}\right) – \left(-8 + \dfrac{8}{3}\right) \)
\( = 16 – \dfrac{16}{3} = \dfrac{32}{3} \)
Final Answer: \( \displaystyle \dfrac{32}{3} \) units²
Finding the Area Between Curves Expressed as Functions of y
When curves are expressed as functions of \( y \) (i.e., \( x = g(y) \) and \( x = h(y) \)), the area between them can be found by integrating with respect to \( y \). This approach is useful when the curves are more easily described as horizontal functions or when vertical slices would require splitting the integral into multiple parts.
General Formula:
If \( g(y) \) is the rightmost function and \( h(y) \) is the leftmost function for \( y \) in the interval \([c, d]\), then the area \( A \) is:
\( A = \displaystyle \int_{c}^{d} \left[ g(y) – h(y) \right] \, dy \)
Steps to Solve:
- Sketch the region to visualize the curves and determine limits of integration in terms of \( y \).
- Identify the rightmost function \( g(y) \) and the leftmost function \( h(y) \) for the given \( y \)-interval.
- Set up the integral \( \displaystyle \int_{c}^{d} \left[ g(y) – h(y) \right] \, dy \).
- Evaluate the definite integral to find the area.
Example :
Find the area of the region bounded by \( x = y^2 \) and \( x = 4 \) between \( y = -2 \) and \( y = 2 \).
▶️ Answer/Explanation
The rightmost function is \( g(y) = 4 \) and the leftmost function is \( h(y) = y^2 \).
Area: \( A = \displaystyle \int_{-2}^{2} \left[ 4 – y^2 \right] \, dy \)
\( A = \left[ 4y – \dfrac{y^3}{3} \right]_{-2}^{2} \)
\( A = \left( 8 – \dfrac{8}{3} \right) – \left( -8 + \dfrac{8}{3} \right) \)
\( A = \dfrac{16}{1} – \dfrac{16}{3} = \dfrac{48 – 16}{3} = \dfrac{32}{3} \)
Example:
Find the area of the region bounded by \( x = \sqrt{y} \) and \( x = y – 2 \) between \( y = 1 \) and \( y = 4 \).
▶️ Answer/Explanation
The rightmost function is \( g(y) = \sqrt{y} \) and the leftmost function is \( h(y) = y – 2 \).
Area: \( A = \displaystyle \int_{1}^{4} \left[ \sqrt{y} – (y – 2) \right] \, dy \)
\( A = \displaystyle \int_{1}^{4} \left[ y^{1/2} – y + 2 \right] \, dy \)
\( A = \left[ \dfrac{2}{3}y^{3/2} – \dfrac{y^2}{2} + 2y \right]_{1}^{4} \)
At \( y = 4 \): \( \dfrac{2}{3}(8) – \dfrac{16}{2} + 8 = \dfrac{16}{3} – 8 + 8 = \dfrac{16}{3} \)
At \( y = 1 \): \( \dfrac{2}{3}(1) – \dfrac{1}{2} + 2 = \dfrac{2}{3} – \dfrac{1}{2} + 2 = \dfrac{2}{3} – \dfrac{1}{2} + \dfrac{4}{2} \)
Difference: \( \dfrac{16}{3} – \left( \dfrac{2}{3} – \dfrac{1}{2} + 2 \right) = \dfrac{16}{3} – \left( \dfrac{2}{3} + \dfrac{3}{2} \right) \)
\( = \dfrac{16}{3} – \dfrac{4}{6} – \dfrac{9}{6} = \dfrac{16}{3} – \dfrac{13}{6} = \dfrac{32}{6} – \dfrac{13}{6} = \dfrac{19}{6} \)