A plate is bounded by the parabola y²=2px and the chord
A plate is bounded by the parabola $y^2=2px$ and the chord through its focus perpendicular to the axis (the latus rectum). Find the mass of the plate if the surface density at each point is inversely proportional to the distance from the directrix.
Explanation of Method
The focus is at $\left(\frac p2,0\right)$ and the directrix is $x=-\frac p2$, so the latus rectum is the vertical chord $x=\frac p2$, and the density is $\mu=\dfrac k{x+p/2}$. Integrating $x$ first (a simple logarithm) leaves a $y$-integral of $\ln\!\left(\dfrac{2p^2}{y^2+p^2}\right)$, which splits into an elementary term and $\int\ln(y^2+p^2)\,dy$, handled by integration by parts.
- Identify the region and the density function.
- Integrate over $x$ to get a $y$-only logarithmic integrand.
- Integrate over $y$, using integration by parts for the $\ln(y^2+p^2)$ piece.
Step 1 · Region and density
The region is $D:\dfrac{y^2}{2p}\le x\le\dfrac p2,\,-p\le y\le p$, and $\mu(x,y)=\dfrac k{x+p/2}$ (distance to the directrix $x=-p/2$).
Step 2 · Integrate over $x$
By symmetry in $y$:
$$ m = 2k\int_0^p\int_{y^2/(2p)}^{p/2}\frac{dx}{x+p/2}\,dy, \qquad \int_{y^2/(2p)}^{p/2}\frac{dx}{x+p/2} = \ln\frac{2p^2}{y^2+p^2}. $$
Step 3 · Integrate over $y$
$$ m = 2k\int_0^p\left[\ln(2p^2)-\ln(y^2+p^2)\right]dy = 2k\left[p\ln(2p^2)-I\right], \qquad I=\int_0^p\ln(y^2+p^2)\,dy. $$
Integrating $I$ by parts ($u=\ln(y^2+p^2),\,dv=dy$):
$$ I = p\ln(2p^2) - 2\int_0^p\frac{y^2}{y^2+p^2}\,dy = p\ln(2p^2) - 2\left(p-\frac{\pi p}4\right) = p\ln(2p^2)-2p+\frac{\pi p}2. $$
Substituting back, the $\ln(2p^2)$ terms cancel:
$$ m = 2k\left[2p-\frac{\pi p}2\right] = kp(4-\pi). $$
Additional Notes
- This derivation matches the source exactly.
- Verified numerically (e.g. $p=1.6,\,k=1$): direct double integration gives $m\approx1.37345$, matching $p(4-\pi)$ exactly.
Result
The mass of the plate is: