Find the area bounded by the following curves: y²=x+2, x=2
Find the area bounded by the following curves: $$ y^2=x+2, \qquad x=2. $$
Explanation of Method
The parabola $y^2=x+2$ opens rightward with vertex at $(-2,0)$; the vertical line $x=2$ cuts it at two symmetric points. Using horizontal strips (integrating in $x$ across each fixed $y$) turns this into a single elementary integral of the gap between the line and the parabola.
- Find the intersection points of the parabola and the line.
- Set up the region using horizontal strips.
- Integrate the horizontal gap from $y=-2$ to $y=2$.
Step 1 · Find the intersection points
Setting $x=2$ in $y^2=x+2$: $y^2=4\Rightarrow y=\pm2$. Intersection points: $(2,2)$ and $(2,-2)$.
Step 2 · Set up the region
Rewriting the parabola as $x=y^2-2$, for each fixed $y\in[-2,2]$ the region runs from the parabola (left boundary, $x=y^2-2$) to the line (right boundary, $x=2$):
$$ A = \int_{-2}^{2}\big[2-(y^2-2)\big]\,dy = \int_{-2}^{2}(4-y^2)\,dy. $$
Step 3 · Evaluate the integral
Since the integrand is even, use symmetry:
$$ A = 2\int_0^{2}(4-y^2)\,dy = 2\left[4y-\frac{y^3}3\right]_0^{2} = 2\left(8-\frac83\right) = 2\cdot\frac{16}3 = \frac{32}3. $$
Additional Notes
- Using vertical strips instead would require splitting the integral (since for $-2\leqslant x\leqslant2$ the region is bounded above and below by the two branches $y=\pm\sqrt{x+2}$ of the same parabola, with no extra split needed here since the line simply caps the region at $x=2$); the horizontal-strip approach avoids any square-root integrands entirely, which is why it was chosen.
Result
The area is: