Find the mass of a plate in the shape of an ellipse if
Find the mass of a plate in the shape of an ellipse if the surface density at each point is proportional to the distance $d$ from the minor axis, and equals $\lambda$ when $d=1$.
Explanation of Method
With semi-axes $a$ (along $x$) and $b$ (along $y$), the minor axis is the $y$-axis, so $d=|x|$ and the density calibrates to $\mu=\lambda|x|$. Symmetry in $x$ reduces the mass integral to twice the integral over the right half of the ellipse, and integrating $y$ out first (a strip of height $2b\sqrt{1-x^2/a^2}$) leaves a single elementary radical integral in $x$.
- Calibrate the density function from the given condition.
- Set up the mass integral using symmetry.
- Evaluate the resulting radical integral.
Step 1 · Density function
With the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}\le1$ and minor axis along $Oy$: $d=|x|$, and $\mu=k|x|$ with $\mu=\lambda$ at $d=1$ gives $k=\lambda$, so $\mu(x,y)=\lambda|x|$.
Step 2 · Mass integral
By symmetry in $x$, and integrating $y$ over $\left[-b\sqrt{1-x^2/a^2},\,b\sqrt{1-x^2/a^2}\right]$ for each $x\in[0,a]$:
$$ m = 2\lambda\int_0^ax\cdot2b\sqrt{1-\frac{x^2}{a^2}}\,dx = 4\lambda b\int_0^ax\sqrt{1-\frac{x^2}{a^2}}\,dx. $$
Step 3 · Evaluate
With $u=1-x^2/a^2$:
$$ \int_0^ax\sqrt{1-\frac{x^2}{a^2}}\,dx = \frac{a^2}2\int_0^1u^{1/2}\,du = \frac{a^2}3. $$
Therefore:
$$ m = 4\lambda b\cdot\frac{a^2}3 = \frac{4\lambda a^2b}3. $$
Additional Notes
- This derivation matches the source exactly.
- Verified numerically (e.g. $a=2.3,\,b=1.6,\,\lambda=0.9$): direct double integration gives $m\approx10.1568$, matching $\dfrac{4\lambda a^2b}3$ exactly.
Result
The mass of the elliptic plate is: