#Study Notes: Area Between a Curve and the Y-Axis

#Table of Contents

1. Introduction
2. Finding the Area Between a Curve and the Y-Axis
3. Worked Example 1
4. No Given Limits
5. Negative Area Integrals
6. Worked Example 2
7. Practice Questions
8. Glossary
9. Summary and Key Takeaways

#Introduction

This section covers the method to find the area between a curve and the y-axis using definite integrals. Understanding this concept is essential for solving problems related to areas under curves in coordinate geometry.

#Finding the Area Between a Curve and the Y-Axis

To find the area between a curve and the y-axis, follow these steps:

#Steps:

1. Identify the function $x = g(y)$.
2. Set the integration limits $y = a$ and $y = b$.
3. Evaluate the definite integral: $\int_{a}^{b} g(y) , dy$.

#Example

Given a function in terms of $x$, $y = f(x)$, rearrange it to $x = g(y)$ before integrating.

#Worked Example 1

#Problem:

Find the area of the region enclosed by the curve $y = 2 + \sqrt{x + 4}$ and the horizontal lines $y = 4$ and $y = 6$.

#Solution:

1. Rearrange the equation: $$\begin{aligned} y & = 2 + \sqrt{x + 4} \\ y - 2 & = \sqrt{x + 4} \\ (y - 2)^2 & = x + 4 \\ x & = (y - 2)^2 - 4 \end{aligned}$$

2. Integrate with respect to $y$ between $y = 4$ and $y = 6$: $$\begin{aligned} \int_{4}^{6} \left[(y - 2)^2 - 4\right] , dy & = \left[ \frac{1}{3}(y - 2)^3 - 4y \right]_{4}^{6} \\ & = \left( \frac{1}{3}(6 - 2)^3 - 4 \cdot 6 \right) - \left( \frac{1}{3}(4 - 2)^3 - 4 \cdot 4 \right) \\ & = \left( \frac{64}{3} - 24 \right) - \left( \frac{8}{3} - 16 \right) \\ & = \left( \frac{64}{3} - 24 \right) - \left( \frac{8}{3} - 16 \right) \\ & = \left( \frac{64 - 72}{3} \right) - \left( \frac{8 - 48}{3} \right) \\ & = \left( \frac{-8}{3} \right) - \left( \frac{-40}{3} \right) \\ & = \frac{32}{3} \end{aligned}$$

The area is $\frac{32}{3}$ square units.

#No Given Limits

#What if limits are not provided?

If the integration limits are not provided, they are often the y-axis intercepts:

1. Find the y-axis intercepts: Solve for $x = 0$.
2. Integrate the function between these intercepts.

The x-axis ($y = 0$) may also be one of the limits.

#Negative Area Integrals

#When is the area integral negative?

• If the area lies to the left of the y-axis, the value of the definite integral will be negative.

  • However, an area cannot be negative.
  • The area is equal to the absolute value of the definite integral.

• For areas that lie both to the right and left of the y-axis:

  • Use the method outlined in the 'Multiple Areas' study guide.

#Worked Example 2

#Problem:

Find the area of the region enclosed by the curve $y = x^2 - 7x + 10$ and the y-axis.

#Solution:

1. Find the y-intercepts by solving for $x = 0$: $$\begin{aligned} 0 & = y^2 - 7y + 10 \\ 0 & = (y - 2)(y - 5) \\ y & = 2 , \text{or} , y = 5 \end{aligned}$$

2. Integrate with respect to $y$ between $y = 2$ and $y = 5$: $$\begin{aligned} \int_{2}^{5} \left(y^2 - 7y + 10\right) , dy & = \left[ \frac{1}{3} y^3 - \frac{7}{2} y^2 + 10y \right]_{2}^{5} \\ & = \left( \frac{1}{3} \cdot 125 - \frac{7}{2} \cdot 25 + 50 \right) - \left( \frac{1}{3} \cdot 8 - \frac{7}{2} \cdot 4 + 20 \right) \\ & = \left( \frac{125}{3} - \frac{175}{2} + 50 \right) - \left( \frac{8}{3} - 14 + 20 \right) \\ & = \left( \frac{250}{6} - \frac{525}{6} + \frac{300}{6} \right) - \left( \frac{16}{6} - \frac{84}{6} + \frac{120}{6} \right) \\ & = \left( \frac{25}{6} - \frac{26}{3} \right) \\ & = -\frac{9}{2} \end{aligned}$$

Since the area cannot be negative, the area is $\frac{9}{2}$ square units.

#Practice Questions

Practice Question

Question 1: Find the area between the curve $x = y^2$ and the y-axis between $y = -1$ and $y = 2$.

Answer: $$\int_{-1}^{2} y^2 , dy = \left[ \frac{1}{3} y^3 \right]_{-1}^{2} = \left( \frac{8}{3} - (-\frac{1}{3}) \right) = \frac{9}{3} = 3 , \text{square units}$$

Practice Question

Question 2: Find the area enclosed by the curve $x = 4 - y^2$ and the y-axis between $y = -2$ and $y = 2$.

Answer: $$\int_{-2}^{2} (4 - y^2) , dy = \left[ 4y - \frac{1}{3} y^3 \right]_{-2}^{2} = \left( 8 - \frac{8}{3} \right) - \left( -8 + \frac{8}{3} \right) = \frac{32}{3} , \text{square units}$$

#Glossary

1. Definite Integral: A type of integral with upper and lower limits, providing the net area under a curve.
2. Y-Axis Intercept: The points where a curve intersects the y-axis.
3. Absolute Value: The non-negative value of a number without regard to its sign.

#Summary and Key Takeaways

#Summary

• The area between a curve and the y-axis can be found using definite integrals.
• Rearrange the function into $x = g(y)$ if needed.
• Integrate between the given limits or find y-axis intercepts if limits are not provided.
• Account for negative areas by taking the absolute value of the definite integral.

#Key Takeaways

• Always ensure the function is in terms of $y$ before integrating.
• Check whether you are finding the value of an integral or an area.
• Remember to take the absolute value for areas to the left of the y-axis.