Find the volume of the part of the cylinder $\frac{x^2}{a^2}+\frac{z^2}{c^2}=1$ which lies between the plane $y=0 ,y=mx$

Question

Find the volume of the part of the cylinder $\frac{x^2}{a^2}+\frac{z^2}{c^2}=1$ which lies between the plane $y=0, y=mx$

The limits of integration which I set went as follows:

$${\int \int}_{D} \int_{y=0}^{mx}\mathrm dy\mathrm dx\mathrm dz$$ where $$D=\big \lbrace \: (x,z) \mid \frac{x^2}{a^2}+\frac{z^2}{c^2}=1 \: \big \rbrace$$

Now let $x=au,z=bv$, then $D$ transforms into $\big \lbrace \: (u,v) \mid u^2 + v^2=1 \: \big \rbrace$

Then the integral becomes $$\int_{u=-1}^1 \int_{v=-\sqrt{1-u^2}}^{v=\sqrt{1-u^2}} ma^2cu\mathrm du\mathrm dv$$

Answer 1

@Antimomy , your approach is absolutely OK. Now just solve the integration.

$$\int_{u=-1}^{u=1}\int_{v=-\sqrt{1-u^2}}^{v=-\sqrt{1-u^2}}ma^2cudvdu=a^2cm\int_{u=-1}^{u=1}\int_{v=-\sqrt{1-u^2}}^{v=-\sqrt{1-u^2}}udvdu$$ In this step, if you change it into the polar form then you will get the answer $0$. Because, see in the picture the same volume is in the negative side also.

First use the properties of definite integral then you can change it into polar. So, $$V=4a^2cm\int_{u=0}^{u=1}\int_{v=0}^{v=-\sqrt{1-u^2}}udvdu$$

Now, put $u=rcos\theta$, $v=rsin\theta$. For positive $u$, $v$ the limits will be, $0\leq r\leq 1$ and $0\leq \theta \leq \frac{\pi}{2}$. $$V=4a^2cm\int_0^{\frac{\pi}{2}}\int_0^1r^2cos\theta drd\theta=\frac{4}{3}a^2cm\int_0^\frac{\pi}{2}cos\theta d\theta=\frac{4}{3}a^2cm.$$

Answer 2

From $ \dfrac{x^2}{a^2} + \dfrac{z^2}{c^2} = 1 $, take

$x = a u \cos t$

$z = c u \sin t$

$y = y$

where $ 0 \le t \le 2 \pi $ and $ 0 \le u \le 1 $

The Jacobian of the above transformation is

$ J = \dfrac{\partial(x, y, z) }{\partial(u, t, y)} = \big| \begin{vmatrix} a \cos t && - a u \sin t && 0 \\ 0 && 0 && 1 \\ c \sin t && c u \cos t && 0 \end{vmatrix} \big| = a c u$

Now the volume is given by the following integral

$\begin{equation} \begin{split} V &= \displaystyle \int_{0}^{2\pi} \int_0^1 \int_{y=0}^{a m u |\cos t|} (a c u) \ dy \ du \ dt =\int_{0}^{2\pi} \int_0^1 (a^2 m c u^2) |\cos t | \ du \ dt \\ &= \dfrac{1}{3} a^2 c m \int_{0}^{2 \pi} | \cos t | dt = \dfrac{4}{3} a^2 c m \end{split} \end{equation}$