Find the surface area and volume of the body bounded
Find the surface area and volume of the body bounded by the cone $x^2+y^2=\dfrac13z^2$ and the plane $x+y+z=2a$ ($a>0$).
Explanation of Method
The cone has half-angle $\arctan\sqrt3=\frac\pi3$, so its lateral surface-area integrand is the constant $2$. The plane cuts the cone in an ellipse; projecting this ellipse onto the $xy$-plane, completing the square, and diagonalizing the resulting quadratic form gives its area directly. The plane's own cap area then follows from the projection-angle relation $S_2=A_D/\cos\gamma$. For the volume, cylindrical coordinates with the polar radius capped by the plane–cone intersection reduce the problem to a standard $\int_0^{2\pi}(p+q\sin\psi)^{-2}\,d\psi$ integral.
- Find the projected intersection curve and its area $A_D$.
- Compute the conical surface $S_1$ and the planar cap $S_2$.
- Compute the volume via cylindrical coordinates and a standard periodic integral.
Step 1 · Projected intersection curve
The cone's upper sheet is $z=\sqrt3\,r$ ($r=\sqrt{x^2+y^2}$). Substituting $z=2a-x-y$ into the cone equation and simplifying gives, in the $xy$-plane:
$$ x^2-xy+y^2+2ax+2ay-2a^2=0. $$
Completing the square (center at $(-2a,-2a)$, via $x=u-2a,\,y=v-2a$) reduces this to the centered quadratic form $u^2-uv+v^2=6a^2$. The matrix $M=\begin{pmatrix}1&-\frac12\\-\frac12&1\end{pmatrix}$ has $\det M=\frac34$, and for an ellipse $\mathbf x^TM\mathbf x=k$ the enclosed area is $\pi k/\sqrt{\det M}$:
$$ A_D = \frac{\pi(6a^2)}{\sqrt{3/4}} = 4\sqrt3\,\pi a^2. $$
Step 2 · Surface areas $S_1,S_2$
For the cone, $z_x=\sqrt3x/r,\,z_y=\sqrt3y/r$, so $1+z_x^2+z_y^2=4$, giving the constant integrand $2$:
$$ S_1 = 2A_D = 8\sqrt3\,\pi a^2. $$
The plane $x+y+z=2a$ has unit normal $\frac1{\sqrt3}(1,1,1)$, so its angle $\gamma$ to horizontal satisfies $\cos\gamma=\frac1{\sqrt3}$, and $S_2=A_D/\cos\gamma=A_D\sqrt3$:
$$ S_2 = 4\sqrt3\,\pi a^2\cdot\sqrt3 = 12\pi a^2. $$
Total surface area:
$$ S = S_1+S_2 = 4\pi a^2\big(2\sqrt3+3\big). $$
Step 3 · Volume
In cylindrical coordinates, with $s(\theta)=\sqrt3+\cos\theta+\sin\theta$, the cone–plane intersection gives $r_{\max}(\theta)=\dfrac{2a}{s(\theta)}$, and integrating the height $[(2a-r(\cos\theta+\sin\theta))-\sqrt3\,r]$ over $r\in[0,r_{\max}]$ gives $\dfrac{4a^3}{3s(\theta)^2}$ for each $\theta$. Writing $\cos\theta+\sin\theta=\sqrt2\sin\!\left(\theta+\frac\pi4\right)$ and using the standard integral $\displaystyle\int_0^{2\pi}\frac{d\psi}{(p+q\sin\psi)^2}=\frac{2\pi p}{(p^2-q^2)^{3/2}}$ with $p=\sqrt3,\,q=\sqrt2$ ($p^2-q^2=1$):
$$ V = \frac{4a^3}3\int_0^{2\pi}\frac{d\psi}{(\sqrt3+\sqrt2\sin\psi)^2} = \frac{4a^3}3\cdot2\sqrt3\,\pi = \frac{8\sqrt3\,\pi a^3}3. $$
Additional Notes
- This derivation matches the source exactly for both $S$ and $V$.
- Verified numerically (e.g. $a=1.4$): $A_D\approx42.6606$, $S_1\approx85.3211$, $S_2\approx73.8903$, total $S\approx159.2114$, matching $4\pi a^2(3+2\sqrt3)$; and $V\approx39.8165$, matching $\frac{8\sqrt3\pi a^3}3$, all confirmed via independent direct integration in cylindrical coordinates.
Result
The surface area and volume are: