Area Between Curves: Functions of y - AP Calculus BC

1. How do you find intersection points algebraically for $x = f(y)$ and $x = g(y)$?

   Set $f(y) = g(y)$ and solve for $y$. Standard method: equate the functions and solve the resulting equation.

2. State the integral setup for curves $x = y^2 + 1$ and $x = 4y - 2$ from $y = -1$ to $y = 3$.

   $\text{integral from } -1 \text{ to } 3 \text{ of } [(4y - 2) - (y^2 + 1)] \text{ dy}$. $(4y - 2)$ is rightmost, so subtract $(y^2 + 1)$ over the given bounds.

3. What is the integral setup for $x = y^2 - 4$ and $x = y + 2$ from $y = -3$ to $y = 3$?

   $\text{integral from } -3 \text{ to } 3 \text{ of } [(y + 2) - (y^2 - 4)] \text{ dy}$. $(y + 2)$ is rightmost, so subtract $(y^2 - 4)$ over the specified interval.

4. Identify the setup to find the area between $x = y^2$ and $x = 4 - y^2$ from $y = -2$ to $y = 2$.

   $\text{Area} = 2 \times \text{integral from } 0 \text{ to } 2 \text{ of } [(4 - y^2) - (y^2)] \text{ dy}$. Uses symmetry since the region is symmetric about the $x$-axis.

5. Find the top function between $x = 2 - y^2$ and $x = y + 1$ for $y$ in $[-1, 2]$.

   $x = y + 1$ is the top function. For any $y$ in the interval, $y + 1$ gives larger $x$-values than $2 - y^2$.

6. What do you need to verify before integrating to find the area between curves?

   The top and bottom functions for each interval. Must identify which function is rightmost to ensure positive area calculation.

7. Determine the area between $x = 4 - y^2$ and $x = y^2 - 4$ for $y \text{ in } [-2, 2]$.

   $\text{Area} = \text{integral from } -2 \text{ to } 2 \text{ of } [(4 - y^2) - (y^2 - 4)] \text{ dy}$. $(4 - y^2)$ is always rightmost since it's positive while $(y^2 - 4)$ is negative.

8. How do you determine which function is the top function?

   Evaluate $f(y)$ and $g(y)$ at points in the interval. Test points in the interval to see which function gives larger $x$-values.

9. What is the integral setup for area between $x = 2y^2$ and $x = y + 3$ from $y = -1$ to $y = 2$?

   $\text{integral from } -1 \text{ to } 2 \text{ of } [(y + 3) - (2y^2)] \text{ dy}$. $(y + 3)$ is rightmost, so subtract $2y^2$ over the specified interval.

10. Identify the integral setup for $x = 2y$ and $x = y^3$ from $y = -2$ to $y = 2$.

    $\text{integral from } -2 \text{ to } 2 \text{ of } [(2y) - (y^3)] \text{ dy}$. $2y$ is rightmost for most of the interval, so subtract $y^3$ from it.

11. What is the significance of intersection points in finding the area between curves?

    They are the limits of integration. Intersection points define where curves meet and serve as integration boundaries.

12. When are horizontal slices preferred over vertical slices?

    When $x$ is expressed as a function of $y$. Horizontal slices avoid complications when $x = f(y)$ is easier to work with.

13. What must be true for $f(y)$ and $g(y)$ to use the area formula between curves?

    Both must be continuous on the interval. Continuity ensures the integral exists and the area formula is valid.

14. What is the first step in finding the area between curves $x = f(y)$ and $x = g(y)$?

    Determine points of intersection. Intersection points determine the integration bounds for the area calculation.

15. Which function is top: $x = y^2 - 2$ or $x = 2y$ for $y$ in $[-1, 2]$?

    $x = 2y$ is the top function. Compare values: at $y = 0$, $2y = 0$ and $y^2 - 2 = -2$, so $2y$ is rightmost.

16. What is the integral setup for the area between $x = y^2 + 4$ and $x = 5y$ from $y = 0$ to $y = 3$?

    $\int_0^3 [(5y) - (y^2 + 4)] \, \mathrm{d}y$. $(5y)$ is rightmost, so subtract $(y^2 + 4)$ over the specified interval.

17. What are the limits of integration for $x = y^2$ and $x = 2y + 3$ from $y = 0$ to $y = 2$?

    $y = 0$ to $y = 2$. The integration bounds are simply the given interval endpoints.

18. Calculate the area between $x = y^2 + 2$ and $x = 3y + 1$ from $y = 0$ to $y = 2$.

    $\int_0^2 [(3y + 1) - (y^2 + 2)] \, dy$. $(3y + 1)$ is rightmost, so subtract $(y^2 + 2)$ over the given bounds.

19. What should you check if the integral yields a negative area?

    Check the order of the functions in the integral. Negative area indicates the order of functions in the integrand was reversed.

20. Which curve should be on top when finding the area between curves expressed as functions of $y$?

    The curve with the greater $y$-value for each $y$ in the interval. The rightmost curve (larger $x$-value) is considered 'on top' for horizontal integration.

21. Identify the integral setup for curves $x = y^2 + 1$ and $x = 2y + 3$ between $y = 0$ and $y = 2$.

    $\text{integral from } 0 \text{ to } 2 \text{ of } [(2y + 3) - (y^2 + 1)] \text{ dy}$. Rightmost function $(2y + 3)$ minus leftmost function $(y^2 + 1)$ over given bounds.

22. State the condition for using vertical slices in finding area between curves.

    Functions must be expressed as $y = f(x)$. Vertical slices integrate with respect to $x$, requiring $y$ as a function of $x$.

23. Identify the top function between $x = 3y$ and $x = y^3$ for $y$ in $[0, 2]$.

    $x = 3y$ is the top function. For positive $y$ values, $3y$ grows linearly while $y^3$ grows more slowly initially.

24. What is the general formula for the area between two curves expressed as functions of $y$?

    $A = \text{integral from } c \text{ to } d \text{ of } [f(y) - g(y)] \text{ dy}$. Standard area formula with horizontal slices where $f(y)$ is the rightmost curve.

25. State the condition for using horizontal slices in finding area between curves.

    Functions must be expressed as $x = f(y)$. Horizontal slices integrate with respect to $y$, requiring $x$ as a function of $y$.

26. Determine the area between $x = y^2 + 1$ and $x = 5y - 3$ from $y = 0$ to $y = 3$.

    $\int_0^3 [(5y - 3) - (y^2 + 1)] \, dy$. $(5y - 3)$ is rightmost, so subtract $(y^2 + 1)$ over the given bounds.

27. State the integral setup for $x = y^3$ and $x = 4y - 1$ from $y = -1$ to $y = 1$.

    $\int_{-1}^{1} [(4y - 1) - y^3] \, dy$. $(4y - 1)$ is rightmost for this interval, so subtract $y^3$ from it.

28. Determine the integral setup for $x = y^2$ and $x = 2 - y$ from $y = -1$ to $y = 1$.

    $\text{integral from } -1 \text{ to } 1 \text{ of } [(2 - y) - (y^2)] \text{ dy}$. $(2 - y)$ is rightmost for this interval, so subtract $y^2$ from it.

29. What happens if $f(y)$ and $g(y)$ switch roles over the interval?

    Split the integral at the intersection point. When curves cross, split the integral at intersection to maintain proper order.

30. Calculate the area between $x = y + 2$ and $x = 2y + 1$ for $y$ in $[1, 3]$.

    $\text{integral from } 1 \text{ to } 3 \text{ of } [(2y + 1) - (y + 2)] \text{ dy}$. $(2y + 1)$ is rightmost, so subtract $(y + 2)$ over the given interval.

31. What is the integral setup for $x = 3y^2$ and $x = y + 4$ from $y = 0$ to $y = 2$?

    $\text{integral from } 0 \text{ to } 2 \text{ of } [(y + 4) - (3y^2)] \text{ dy}$. $(y + 4)$ is rightmost, so subtract $3y^2$ over the specified interval.

32. Identify the area calculation between $x = y + 2$ and $x = 3y + 1$ for $y$ in $[0, 2]$.

    $\text{integral from } 0 \text{ to } 2 \text{ of } [(3y + 1) - (y + 2)] \text{ dy}$. $(3y + 1)$ is rightmost, so subtract $(y + 2)$ over the given bounds.

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