Find the surface area of the body bounded by the surfaces

Find the surface area of the body bounded by the surfaces $x^2+y^2=2az$, $(x^2+y^2)^2+2a^2xy=0$, and $z=0$.

Step 1 · The bounding curve

In polar coordinates: $r^4+2a^2r^2\cos\theta\sin\theta=0\iff r^2\big(r^2+a^2\sin2\theta\big)=0$, so for $r\ne0$: $r^2=-a^2\sin2\theta$, real for $\theta\in\left[\frac\pi2,\pi\right]\cup\left[\frac{3\pi}2,2\pi\right]$ (two lobes).

Step 2 · Base area $S_2$

By symmetry the two lobes contribute equally:

$$ S_2 = 2\cdot\frac12\int_{\pi/2}^\pi\big(-a^2\sin2\theta\big)\,d\theta = a^2\left[\frac{\cos2\theta}2\right]_{\pi/2}^{\pi} = a^2\left(\frac12-\left(-\frac12\right)\right) = a^2. $$

Step 3 · Paraboloid cap $S_1$

With $z=\dfrac{r^2}{2a}$, the integrand is $\sqrt{1+r^2/a^2}$. The inner radial integral (via $u=1+r^2/a^2$, using $r_{\max}^2/a^2=-\sin2\theta$) gives:

$$ \int_0^{r_{\max}}r\sqrt{1+\frac{r^2}{a^2}}\,dr = \frac{a^2}3\Big[(1-\sin2\theta)^{3/2}-1\Big], $$

so:

$$ S_1 = \frac{2a^2}3\int_{\pi/2}^\pi\Big[(1-\sin2\theta)^{3/2}-1\Big]\,d\theta. $$

Substituting $\varphi=2\theta-\pi$ (so $\sin2\theta=-\sin\varphi$) and then $1+\sin\varphi=2\cos^2\!\left(\frac\pi4-\frac\varphi2\right)$ reduces this to an elementary power of cosine, giving $\displaystyle\int_0^\pi(1+\sin\varphi)^{3/2}\,d\varphi=\frac{20}3$, so:

$$ S_1 = \frac{a^2}3\left(\frac{20}3-\pi\right) = \frac{a^2(20-3\pi)}9. $$

Step 4 · Total

$$ S = S_1+S_2 = \frac{a^2(20-3\pi)}9 + a^2 = \frac{a^2(29-3\pi)}9. $$

Additional Notes

• A note on the source computation: the problem statement as originally given uses $(x^2+y^2)^2+2axy=0$ for the lateral cylinder (linear in $a$), which is dimensionally inconsistent with the paraboloid $x^2+y^2=2az$: in the paraboloid, $a$ carries units of length (so that $2az\sim L^2$), while in the cylinder as literally written, $a$ would need to carry units of area (so that $2a\,xy\sim L^4$ matches $(x^2+y^2)^2\sim L^4$). Reading the cylinder with the same length-scale $a$ throughout forces the equation $(x^2+y^2)^2+2a^2xy=0$, as used above; this is the reading adopted here, and it is also the one implicitly used by the source itself — the source's own working sets "$a=1$" partway through specifically to sidestep this inconsistency, then "restores" $a^2$ (not $a$) at the very end by an unexplained appeal to "dimensional analysis." Verified numerically across several values of $a$: the boxed formula $S=\frac{a^2(29-3\pi)}9$ holds exactly under the $a^2$ reading of the cylinder for every $a$ tested, while it only accidentally matches the literal $2axy$ reading at $a=1$ (where $a=a^2$) and is otherwise wrong for that reading.
• With this reading adopted, every remaining step (the trigonometric reduction to $\int_0^\pi(1+\sin\varphi)^{3/2}\,d\varphi=\frac{20}3$, and the final assembly) was verified independently and found fully correct.