Calculate the volume of intersection of $x^2+y^2+z^2=4$ and $r=2\cos \theta$ by using cylindrical coordinates.

Question

Calculate the volume of intersection of $x^2+y^2+z^2=4$ and $r=2\cos \theta$ by using cylindrical coordinates.

My try:Intersection will be a cylinder

$x^2+y^2+z^2=4\implies r^2+z^2=4$

Then $$V=\int_0^4\int_0^{2\pi}\int_0^2 rdrd\theta dz$$

My problem:Are the limits of $r,\theta,z $ correct?I just need formula.

Thanks.

Answer 1

Not quite. Imaging slicing the solid at some fixed $z$, treating $z$ as a constant. Then we have two curves in this cross-section: the circle centred at the origin with radius $\sqrt{4 - z^2}$ (namely, $r = \sqrt{4 - z^2}$) and the circle centred at $(1,0)$ with radius $1$ (namely, $r = 2\cos\theta$). Assuming that $-2 < z < 2$, notice that these two curves will intersect in two places. We solve a system of equations to find the value of $\theta$ in terms of $z$ at these two intersection points: $$ 2\cos\theta = r = \sqrt{4 - z^2} \iff \cos\theta = \left(\frac{1}{2}\sqrt{4 - z^2}\right) $$ Now since $z \in (-2, 2)$, notice that the RHS is in $(0, 1)$. Hence, we obtain two solutions for $\theta$ in the interval $(-\pi/2, \pi/2)$, namely: $$ \theta = \pm\arccos\left(\frac{\sqrt{4 - z^2}}{2}\right) $$ Hence, we obtain: $$ V=\int_{-2}^2\int_{-\arccos\left(\frac{1}{2}\sqrt{4 - z^2}\right)}^{\arccos\left(\frac{1}{2}\sqrt{4 - z^2}\right)}\int_{2\cos\theta}^{\sqrt{4 - z^2}} r \, dr \, d\theta \, dz $$

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Now that's a pretty ugly integral. So let's try integrating in a different order: $z \to r \to \theta$. Now imagine slicing the solid at some fixed $\theta$, treating $\theta$ as a constant. Think of drawing this slice in a new "$rz$-half-plane" so that $z$ is a function of $r$ and $r \geq 0$. Then this cross-section has two curves: the circle centred at the origin with radius $2$ (namely, $r^2 + z^2 = 4$) and the vertical line $r = 2\cos\theta$. Assuming that $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$, notice that these curves will intersect in two places, so that the desired region is bounded by four curves, namely:

• $z = \sqrt{4 - r^2}$
• $r = 2\cos\theta$
• $z = -\sqrt{4 - r^2}$
• $r = 0$

Hence, we obtain: $$ V = \int_{-\pi/2}^{\pi/2} \int_0^{2\cos\theta} \int_{-\sqrt{4 - r^2}}^{\sqrt{4 - r^2}} r \, dz \, dr \, d\theta $$ which is perhaps slightly easier to compute by hand.