Find the area bounded by the following curves: y=sqrt x
Find the area bounded by the following curves: $$ y=\sqrt x, \qquad y=2\sqrt x, \qquad y=4. $$
Explanation of Method
Rewriting both curves as $x$ in terms of $y$ ($x=y^2$ and $x=\frac{y^2}4$ respectively) turns this into a single horizontal-strip integral: for each $y$ between $0$ and $4$, the region runs from the curve $y=2\sqrt x$ (on the left) to $y=\sqrt x$ (on the right), avoiding the need to split the integral at the curves' intersection with $y=4$.
- Find where each curve meets $y=4$.
- Rewrite both curves as functions $x(y)$ and identify which is to the left.
- Integrate the horizontal gap between the curves from $y=0$ to $y=4$.
Step 1 · Locate the boundary intersections
$y=\sqrt x$ meets $y=4$ at $x=16$; $y=2\sqrt x$ meets $y=4$ at $x=4$; the two curves $y=\sqrt x$ and $y=2\sqrt x$ meet only at the origin (for $x>0$).
Step 2 · Rewrite as functions of $y$
From $y=\sqrt x$: $x=y^2$. From $y=2\sqrt x$: $x=\dfrac{y^2}4$. Since $2\sqrt x>\sqrt x$ for $x>0$, the curve $y=2\sqrt x$ sits to the left of $y=\sqrt x$ at each height $y$, i.e. $\dfrac{y^2}4<y^2$.
Step 3 · Set up and evaluate the area integral
For $0\leqslant y\leqslant4$, the region runs from $x=\dfrac{y^2}4$ to $x=y^2$:
$$ A = \int_0^4\left(y^2-\frac{y^2}4\right)dy = \frac34\int_0^4 y^2\,dy = \frac34\left[\frac{y^3}3\right]_0^4 = \frac34\cdot\frac{64}3 = 16. $$
Additional Notes
- Verified by the vertical-strip approach: splitting at $x=4$ (where the upper boundary switches from $y=2\sqrt x$ to $y=4$), $\displaystyle\int_0^4(4-2\sqrt x)\,dx+\int_4^{16}(4-\sqrt x)\,dx=\frac{16}3+\frac{32}3=16$, matching the direct computation and confirming the choice of horizontal strips avoided an unnecessary split.
Result
The area is: