Double Integral 157
Medium Polar Coordinates and Change of Variables

Using polar coordinates, find the volume of the solid

Using polar coordinates, find the volume of the solid bounded by the paraboloid: $$ z=x^2+y^2, $$ the cylinders: $$ x^2+y^2=x,\qquad x^2+y^2=2x, $$ and the plane $z=0$.

Explanation of Method

Both cylinders are circles through the origin centered on the positive $x$-axis, with polar equations $r=\cos\theta$ and $r=2\cos\theta$. The base region is the annular-shaped strip between them, and the paraboloid's height $z=r^2$ makes the volume integral straightforward in polar form.

  1. Convert both cylinders to polar form.
  2. Set up the volume integral over the strip $\cos\theta\leqslant r\leqslant2\cos\theta$.
  3. Evaluate using the reduction formula for $\int\cos^4\theta\,d\theta$.

Step 1 · Convert the cylinders to polar form

$x^2+y^2=x\Rightarrow r=\cos\theta$; $x^2+y^2=2x\Rightarrow r=2\cos\theta$. Both require $\cos\theta\geqslant0$, i.e. $\theta\in\left[-\frac\pi2,\frac\pi2\right]$.

Step 2 · Set up the volume integral

$$ V = \int_{-\pi/2}^{\pi/2}\!\int_{\cos\theta}^{2\cos\theta}r^2\cdot r\,dr\,d\theta = \int_{-\pi/2}^{\pi/2}\!\int_{\cos\theta}^{2\cos\theta}r^3\,dr\,d\theta. $$

Step 3 · Evaluate the inner integral

$$ \int_{\cos\theta}^{2\cos\theta}r^3\,dr = \frac{(2\cos\theta)^4-\cos^4\theta}4 = \frac{15\cos^4\theta}4. $$

Step 4 · Evaluate the outer integral

Using $\cos^4\theta=\dfrac{3+4\cos2\theta+\cos4\theta}8$:

$$ \int_{-\pi/2}^{\pi/2}\cos^4\theta\,d\theta = \frac18\left[3\theta+2\sin2\theta+\frac{\sin4\theta}4\right]_{-\pi/2}^{\pi/2} = \frac18\cdot3\pi = \frac{3\pi}8. $$

$$ V = \frac{15}4\cdot\frac{3\pi}8 = \frac{45\pi}{32}. $$


Additional Notes

Result

The volume is:

$$ V = \frac{45\pi}{32}. $$