Find the area bounded by the curve x = 6 - y - y²

Find the area bounded by the curve $x = 6 - y - y^2$ and the $y$-axis.

We will evaluate the area by integrating with respect to $y$.

Step 1: Find intersection points with the $y$-axis

The $y$-axis corresponds to $x = 0$.

Setting $x = 0$:

\[0 = 6 - y - y^2\]

\[y^2 + y - 6 = 0\]

Factoring:

\[(y + 3)(y - 2) = 0\]

Therefore: $y = -3$ or $y = 2$

The curve intersects the $y$-axis at $y = -3$ and $y = 2$.

Step 2: Determine which side of the $y$-axis the curve is on

For $y \in (-3, 2)$, let's check the sign of $x = 6 - y - y^2$.

At $y = 0$:

\[x = 6 - 0 - 0 = 6 > 0\]

So the curve is to the right of the $y$-axis (in the region $x > 0$).

Let's verify at another point, $y = 1$:

\[x = 6 - 1 - 1 = 4 > 0 \quad \checkmark\]

Step 3: Set up the integral

Since the curve is given as $x = f(y)$, we integrate with respect to $y$:

\[A = \int_{-3}^{2} x \, dy = \int_{-3}^{2} (6 - y - y^2) \, dy\]

Step 4: Calculate the antiderivative

\[\int (6 - y - y^2) \, dy = 6y - \frac{y^2}{2} - \frac{y^3}{3} + C\]

Step 5: Evaluate the definite integral

\[A = \left[6y - \frac{y^2}{2} - \frac{y^3}{3}\right]_{-3}^{2}\]

At $y = 2$:

\[6(2) - \frac{2^2}{2} - \frac{2^3}{3} = 12 - 2 - \frac{8}{3} = 10 - \frac{8}{3} = \frac{30 - 8}{3} = \frac{22}{3}\]

At $y = -3$:

\[6(-3) - \frac{(-3)^2}{2} - \frac{(-3)^3}{3} = -18 - \frac{9}{2} - \frac{-27}{3}\]

\[= -18 - \frac{9}{2} + 9 = -9 - \frac{9}{2} = -\frac{18 + 9}{2} = -\frac{27}{2}\]

Therefore:

\[A = \frac{22}{3} - \left(-\frac{27}{2}\right) = \frac{22}{3} + \frac{27}{2}\]

Finding common denominator:

\[A = \frac{44}{6} + \frac{81}{6} = \frac{125}{6}\]

\[\boxed{A = \frac{125}{6} \text{ square units}}\]

Step 6: Numerical value

\[A = \frac{125}{6} \approx 20.833 \text{ square units}\]

Verification

Let's verify the calculation at $y = -3$:

\[6(-3) - \frac{9}{2} + 9 = -18 - 4.5 + 9 = -13.5 = -\frac{27}{2} \quad \checkmark\]

And:

\[\frac{22}{3} + \frac{27}{2} = \frac{44 + 81}{6} = \frac{125}{6} \quad \checkmark\]

Summary

The area bounded by the curve $x = 6 - y - y^2$ and the $y$-axis is:

\[\boxed{A = \frac{125}{6} \approx 20.833 \text{ square units}}\]

Key steps:

1. Find intersections with $y$-axis: $y^2 + y - 6 = 0 \implies y = -3, 2$
2. Verify curve is to the right of $y$-axis for $y \in (-3, 2)$
3. Set up integral: $A = \int_{-3}^{2} (6 - y - y^2) \, dy$
4. Find antiderivative: $6y - \frac{y^2}{2} - \frac{y^3}{3}$
5. Evaluate: $\frac{22}{3} - (-\frac{27}{2}) = \frac{125}{6}$

Key insight:

When the curve is given as $x = f(y)$, integrate with respect to $y$ along the $y$-axis bounds.