Question

The area of the region in the first quadrant that is bounded by the parabola

Hint:

When curves are easier to express as \(x=f(y)\), integrate with respect to \(y\) using \[ \text{Area}=\int (x_{\text{right}}-x_{\text{left}})\,dy. \]

Answer 1

Correct Answer: 54

Step 1: Write curves in terms of $y$.
From $y^2=36x$, $x=\frac{y^2}{36}$; from $y=3x-9$, $x=\frac{y+9}3$.

Step 2: Find the intersection.
Substituting gives $y^2-12y-108=0$, so $y=18$ (first quadrant) at $(9,18)$.

Step 3: Integrate in $y$.
For $0\le y\le18$ the line is to the right, so area $=\int_0^{18}\left(\frac{y+9}3-\frac{y^2}{36}\right)dy$.

Step 4: Evaluate.
The three pieces give $54+54-54=54$.

Step 5: Conclude.
\[ \boxed{54} \]