To find the area of the region bounded by the curve $x = y^2$ and the line $x = y + 6$ between $y = 0$ and $y = 3$, we proceed as follows:

Step 1: Set up the integral

The area is determined by integrating the difference between the rightmost function ($x = y+6$) and the leftmost function ($x = y^2$) over the interval $y \in [0, 3]$.

The area $A$ is: $A = \int_0^3 \left[(y+6) - (y^2)\right] \, dy$

Step 2: Simplify the integrand

Simplify the expression inside the integral: $(y+6) - (y^2) = -y^2 + y + 6$

So the integral becomes: $A = \int_0^3 (-y^2 + y + 6) \, dy$

Step 3: Integrate

We now compute the integral: $\int (-y^2 + y + 6) \, dy = \int -y^2 \, dy + \int y \, dy + \int 6 \, dy$

Compute each term:

1. $\int -y^2 \, dy = -\frac{y^3}{3}$,
2. $\int y \, dy = \frac{y^2}{2}$,
3. $\int 6 \, dy = 6y$.

Thus: $\int (-y^2 + y + 6) \, dy = -\frac{y^3}{3} + \frac{y^2}{2} + 6y$

Step 4: Evaluate the definite integral

Now evaluate this from $y = 0$ to $y = 3$: $A = \left[-\frac{y^3}{3} + \frac{y^2}{2} + 6y\right]_0^3$

At $y = 3$: $-\frac{(3)^3}{3} + \frac{(3)^2}{2} + 6(3) = -\frac{27}{3} + \frac{9}{2} + 18 = -9 + \frac{9}{2} + 18 = 9 + \frac{9}{2} = \frac{18}{2} + \frac{9}{2} = \frac{27}{2}$

At $y = 0$: $-\frac{(0)^3}{3} + \frac{(0)^2}{2} + 6(0) = 0$

So the total area is: $A = \frac{27}{2} - 0 = \frac{27}{2}$

Final Answer:

$\boxed{\frac{27}{2} \, \text{square units}}$

Would you like to see a visual explanation or further steps?

Expanding on this:

1. What is the significance of setting up the integral with respect to $y$ instead of $x$?
2. How can the intersection points of the curves be determined if not given?
3. Can this problem be solved by switching to integration with respect to $x$? How?
4. What role does symmetry play in problems like these?
5. How do the bounds of integration affect the final result?

Tip:

Always sketch the region to confirm the boundaries and understand which function lies to the right or above for a correct setup of the integral.