Find the volume of the solid bounded by the cone
Find the volume of the solid bounded by the cone: $$ z^2=a^2\big(x^2+y^2\big) $$ and the surface: $$ x^2+y^2-bx=b\sqrt{x^2+y^2}, $$ where $a,b>0$ and $z>0$.
Explanation of Method
With $z>0$, the cone gives height $z=ar$ (in polar coordinates). The second equation, in polar form, is the classic cardioid $r=b(1+\cos\theta)$, so the volume is an integral of $ar$ over the cardioid's interior.
- Convert both surfaces to polar coordinates.
- Set up the volume integral over the cardioid.
- Evaluate, expanding $(1+\cos\theta)^3$.
Step 1 · Convert to polar coordinates
Cone: $z=a\sqrt{x^2+y^2}=ar$. Boundary curve: $r^2-br\cos\theta=br\Rightarrow r=b(1+\cos\theta)$ (for $r>0$), the cardioid.
Step 2 · Set up the volume integral
$$ V = \int_0^{2\pi}\!\int_0^{b(1+\cos\theta)}ar\cdot r\,dr\,d\theta = \frac{ab^3}3\int_0^{2\pi}(1+\cos\theta)^3\,d\theta. $$
Step 3 · Expand and integrate
$(1+\cos\theta)^3=1+3\cos\theta+3\cos^2\theta+\cos^3\theta$. Term by term over $[0,2\pi]$: $\displaystyle\int1\,d\theta=2\pi$; $\displaystyle\int\cos\theta\,d\theta=0$; $\displaystyle\int\cos^2\theta\,d\theta=\pi$; $\displaystyle\int\cos^3\theta\,d\theta=0$ (each odd-power cosine integral vanishes over a full period). So:
$$ \int_0^{2\pi}(1+\cos\theta)^3\,d\theta = 2\pi+0+3\pi+0 = 5\pi. $$
Step 4 · Final result
$$ V = \frac{ab^3}3\cdot5\pi = \frac{5\pi ab^3}3. $$
Additional Notes
- This derivation matches the source exactly.
- Verified symbolically: $\displaystyle\int_0^{2\pi}(1+\cos\theta)^3\,d\theta=5\pi$ exactly, giving $V=\dfrac{5\pi ab^3}3$.
Result
The volume is: