Find the area bounded by the parabolas x² = 2py and y² =

Find the area bounded by the parabolas $x^2 = 2py$ and $y^2 = 2px$.

We will find where the two parabolas cross and integrate the region between them.

Step 1: Understand the parabolas

First parabola: $x^2 = 2py \implies y = \frac{x^2}{2p}$. This opens upward with vertex at the origin.

Second parabola: $y^2 = 2px \implies x = \frac{y^2}{2p}$, equivalently $y = \sqrt{2px}$ for $y \geq 0$. This opens to the right with vertex at the origin.

Both parabolas pass through the origin.

Step 2: Find intersection points

From the second: $x = \frac{y^2}{2p}$. Substituting into the first:

\[\left(\frac{y^2}{2p}\right)^2 = 2py\]

\[\frac{y^4}{4p^2} = 2py\]

\[y^4 = 8p^3 y\]

\[y^4 - 8p^3 y = 0\]

\[y(y^3 - 8p^3) = 0\]

So $y = 0$ or $y^3 = 8p^3$, giving $y = 2p$.

When $y = 0$: $x = 0$, point $(0, 0)$

When $y = 2p$:

\[x^2 = 2p(2p) = 4p^2 \implies x = 2p \quad \text{(taking positive root)}\]

Point: $(2p, 2p)$. The parabolas intersect at $(0, 0)$ and $(2p, 2p)$.

Step 3: Determine which parabola is above

For $x \in (0, 2p)$, let's check at $x = p$:

First parabola: $y = \frac{p^2}{2p} = \frac{p}{2}$

Second parabola: $y^2 = 2p \cdot p = 2p^2 \implies y = \sqrt{2}p \approx 1.414p$

So the second parabola ($y^2 = 2px$, i.e. $y = \sqrt{2px}$) is above the first ($x^2 = 2py$).

Step 4: Set up the integral

We integrate with respect to $x$ from $0$ to $2p$:

\[A = \int_0^{2p} \left[\sqrt{2px} - \frac{x^2}{2p}\right] dx\]

Step 5: Evaluate the integral

\[A = \int_0^{2p} \sqrt{2px} \, dx - \int_0^{2p} \frac{x^2}{2p} \, dx\]

First integral:

\[\int_0^{2p} \sqrt{2px} \, dx = \sqrt{2p} \int_0^{2p} x^{1/2} \, dx = \sqrt{2p} \left[\frac{2x^{3/2}}{3}\right]_0^{2p}\]

\[= \sqrt{2p} \cdot \frac{2}{3}(2p)^{3/2} = \sqrt{2p} \cdot \frac{2}{3} \cdot 2\sqrt{2}p^{3/2}\]

\[= \frac{2 \cdot 2\sqrt{2}}{3} \cdot \sqrt{2p} \cdot p^{3/2} = \frac{4\sqrt{2}}{3} \cdot \sqrt{2} \cdot p^2 = \frac{8p^2}{3}\]

Second integral:

\[\int_0^{2p} \frac{x^2}{2p} \, dx = \frac{1}{2p}\left[\frac{x^3}{3}\right]_0^{2p} = \frac{1}{2p} \cdot \frac{(2p)^3}{3} = \frac{1}{2p} \cdot \frac{8p^3}{3} = \frac{4p^2}{3}\]

Step 6: Calculate the area

\[A = \frac{8p^2}{3} - \frac{4p^2}{3} = \frac{4p^2}{3}\]

\[\boxed{A = \frac{4p^2}{3} \text{ square units}}\]

Step 7: Verification

For $p = 1$: $A = \frac{4}{3} \approx 1.333$ square units. The parabolas are $x^2 = 2y$ and $y^2 = 2x$, intersecting at $(0, 0)$ and $(2, 2)$.

Alternative: Integration with respect to $y$

By the symmetry of the two parabolas under swapping $x \leftrightarrow y$ (which maps one parabola onto the other), integrating with respect to $y$ instead gives:

\[A = \int_0^{2p} \left[\sqrt{2py} - \frac{y^2}{2p}\right] dy = \frac{4p^2}{3}\]

the same result, as expected. ✓

Summary

The area bounded by the parabolas $x^2 = 2py$ and $y^2 = 2px$ is:

\[\boxed{A = \frac{4p^2}{3} \text{ square units}}\]

Key steps:

1. Find intersections: $(0, 0)$ and $(2p, 2p)$
2. Determine upper curve: $y = \sqrt{2px}$ (from $y^2 = 2px$)
3. Set up integral: $A = \int_0^{2p} \left[\sqrt{2px} - \frac{x^2}{2p}\right] dx$
4. Evaluate: $\frac{8p^2}{3} - \frac{4p^2}{3} = \frac{4p^2}{3}$

Key insight:

The two parabolas are symmetric with respect to the line $y = x$, and they enclose a lens-shaped region between the origin and the point $(2p, 2p)$.