Using polar coordinates, find the volume of the solid
Using polar coordinates, find the volume of the solid bounded by the planes: $$ z=ax,\qquad z=0 $$ and the cylinder: $$ x^2+y^2=2ax. $$
Explanation of Method
The cylinder $x^2+y^2=2ax$ is a circle of radius $a$ centered at $(a,0)$, with polar equation $r=2a\cos\theta$. The inclined plane $z=ax=ar\cos\theta$ gives the height, non-negative exactly where $\cos\theta\geqslant0$.
- Convert the cylinder to polar form.
- Set up the volume integral with height $z=ar\cos\theta$.
- Evaluate, reusing the $\int\cos^4\theta\,d\theta$ result from Problem 157.
Step 1 · Polar form of the cylinder
$x^2+y^2=2ax\Rightarrow r=2a\cos\theta$, requiring $\theta\in\left[-\frac\pi2,\frac\pi2\right]$ (where $z=ax\geqslant0$ as well, consistently).
Step 2 · Set up the volume integral
$$ V = \int_{-\pi/2}^{\pi/2}\!\int_0^{2a\cos\theta}(ar\cos\theta)\,r\,dr\,d\theta = a\int_{-\pi/2}^{\pi/2}\cos\theta\left[\int_0^{2a\cos\theta}r^2\,dr\right]d\theta. $$
Step 3 · Evaluate the inner integral
$$ \int_0^{2a\cos\theta}r^2\,dr = \frac{(2a\cos\theta)^3}3 = \frac{8a^3\cos^3\theta}3. $$
Step 4 · Evaluate the outer integral
$$ V = \frac{8a^4}3\int_{-\pi/2}^{\pi/2}\cos^4\theta\,d\theta = \frac{8a^4}3\cdot\frac{3\pi}8 = \pi a^4. $$
Additional Notes
- This derivation matches the source exactly.
- Verified symbolically: direct evaluation of the nested integral gives exactly $\pi a^4$.
Result
The volume is: