Find the volume of the solid bounded above by a surface z
Find the volume of the solid bounded above by a surface $z$ over the region enclosed by: $$ x^2+y^2=cx\qquad\text{and}\qquad x^4+y^4=a^2\big(x^2+y^2\big), $$ with $z=0$ below.
Explanation of Method
A note on this problem: as pasted, the problem statement gives two bounding curves in the $xy$-plane but never specifies the height function $z$ that caps the solid from above — a genuine omission, not a transcription slip (there is no natural single "$z=\ldots$" that fits alongside the two given curves). Following the most natural completion — and the one used in the source itself — this page assumes the paraboloid $z=x^2+y^2$ as the height function, and takes the base region to be the disk $x^2+y^2\leqslant cx$ (the circle, which lies entirely inside the quartic curve for reasonable parameter ranges, making it the natural bounding curve for a $z=x^2+y^2$ volume; the second curve would matter for a different, unspecified height function).
- Convert the circular boundary to polar form.
- Set up the volume integral over the disk, using the assumed height $z=x^2+y^2$.
- Evaluate using the standard $\cos^4\theta$ reduction.
Step 1 · Polar form of the circular boundary
$x^2+y^2=cx\Rightarrow r=c\cos\theta$ (for $r>0$, requiring $\cos\theta>0$, i.e. $\theta\in\left(-\frac\pi2,\frac\pi2\right)$).
Step 2 · Set up the volume integral
With the assumed height $z=x^2+y^2=r^2$:
$$ V=\int_{-\pi/2}^{\pi/2}\int_0^{c\cos\theta}r^2\cdot r\,dr\,d\theta = \frac{c^4}4\int_{-\pi/2}^{\pi/2}\cos^4\theta\,d\theta. $$
Step 3 · Evaluate
Using $\displaystyle\int_{-\pi/2}^{\pi/2}\cos^4\theta\,d\theta=\frac{3\pi}8$:
$$ V = \frac{c^4}4\cdot\frac{3\pi}8 = \frac{3\pi c^4}{32}. $$
Additional Notes
- A note on the source: the source itself explicitly acknowledged the missing height function ("the problem statement may be incomplete regarding the top surface"), and adopted the same assumption ($z=x^2+y^2$ over the circular region) used here. Under that assumption, the arithmetic is confirmed correct by direct numerical integration.
- The second curve, $x^4+y^4=a^2(x^2+y^2)$, plays no role in this reading of the problem; it would only become relevant for a differently-specified height function or a differently-bounded region (e.g. the region between the two curves), which the available problem statement does not clarify. If the intended problem is different from this reading, the volume would need to be recomputed accordingly.
Result
Under the assumption $z=x^2+y^2$ over the disk $x^2+y^2\leqslant cx$, the volume is: