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Find the volume bounded by the cylinder $x^2 + y^2=1$ and the planes $y=z , x=0 ,z=0$ in the first octant.

How do I go about doing this?

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  • $\begingroup$ no comment dadu Americo. $\endgroup$ – krishan acton Feb 19 '14 at 17:21
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In general the volume of a region $S$ bounded above by a surface defined by a function of two variables $z=f(x,y)\ge 0$, whose domain is $R$, and below by the plane $z=0$, is given by:

  • the double integral $\iint_R f(x,y)\,dA=\iint_R f(x,y)\,dx\,dy$,
  • or by the triple integral $\iiint_S \,dV=\iiint_S \,dx\,dy\,dz.$

The following figure represents the given region of the first octant ($x\ge 0,y\ge 0,z\ge 0$) bounded by the cylinder $x^{2}+y^{2}=1$ (gray) and the plane $y=z$ (green).

enter image description here

The domain $R$ of the function $z=f(y)=y$ is defined in Cartesian coordinates $x,y$ by

\begin{eqnarray*} R &=&\left\{ (x,y)\in\mathbb{R}^{2}:0\leq x^{2}+y^{2}\leq 1,x\ge 0,y\ge 0\right\} . \end{eqnarray*}

or in polar coordinates $r,\theta$, with $x=r\cos \theta , y=r\sin \theta $, by

\begin{eqnarray*} T &=&\left\{ (r,\theta )\in \mathbb{R}^{2}:0\leq r\leq 1,0\leq \theta \leq \pi /2\right\} . \end{eqnarray*}

As such the requested volume $V$ may be given in Cartesian coordinates by the double integral

\begin{equation*} V=\iint_{R}z\,dx\,dy=\iint_{R}y\,dx\,dy. \end{equation*}

Using polar coordinates, since the Jacobian determinant of the transformation from Cartesian to polar coordinates is $J=\frac{\partial (x,y)}{\partial (r,\theta )}=r$, the integral is converted into the following separable one:

\begin{eqnarray*} V &=&\iint_{T}(r\sin \theta)\, |J| \,dr\,d\theta =\int_{r=0}^{1}\int_{\theta =0}^{\pi /2}r^{2}\sin \theta \,dr\,d\theta \\ &=&\left( \int_{0}^{1}r^{2}dr\right) \left( \int_{0}^{\pi /2}\sin \theta \,d\theta \right) =\left. \frac{1}{3}r^{3}\right\vert _{0}^{1}\left. \left( -\cos \theta \right) \right\vert_{0}^{\pi /2} =\frac{1}{3}. \end{eqnarray*}

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This might be easier in cylindrical coordinates. You just want the region in which $r$ runs from $0$ to $1$, $\theta$ runs from $0$ to $\frac{\pi}2$, and $z$ runs from $0$ to $y=r\sin\theta$. That gives:

$$\int_0^{\frac\pi2}\int_0^1\int_0^{r\sin\theta}r\,dz\,dr\,d\theta$$,

Which I believe comes out to $\frac13$.

The hard part of these things is visualizing the shape and setting up the integral. Do you see where the limits on each of those variables comes from?

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  • $\begingroup$ How do you convert it to cylindrical coordinates? $\endgroup$ – Edgar Simmons Feb 17 '14 at 17:32
  • $\begingroup$ Just use $x=r\cos\theta$ and $y=r\sin\theta$. $\endgroup$ – G Tony Jacobs Feb 17 '14 at 18:05
  • $\begingroup$ Thanks, I have worked it out now. $\endgroup$ – Edgar Simmons Feb 17 '14 at 18:18

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