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Find volume of $E$ using triple integration and cylindrical coordinates, when $E$ is bounded by $$x^2+y^2=x,\quad y=0,\quad y=x,\quad z=0,\quad z=\sqrt{x^2+y^2}$$ I know that in cylindrical coordinates $$x=r\cos\varphi,\quad y=r\sin\varphi,\quad z=h,\quad \text{ where } r\geq0 \text{ and } 0\leq\varphi\leq2\pi$$ but I'm very confused how to set up this integral. Cylindrical coordinates are quite new for me and it's hard to understand, how to make this conversion. So I would be grateful if anyone can help me with this.

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$x^2+y^2=x$ in polar form is just $r=\cos\varphi$. The lines $y=0$ and $y=x$ intersect this curve at $\varphi=0$ and $\varphi=\frac\pi4$ respectively. So our outer integrals are $$\int_{\varphi=0}^{\pi/4}\int_{r=0}^{\cos\varphi}r\,dr\,d\varphi$$ The bounds of $z$ are just $z=0$ to $z=r$ in cylindrical coordinates, so our final answer is $$\int_{\varphi=0}^{\pi/4}\int_{r=0}^{\cos\varphi}\int_{z=0}^rr\,dz\,dr\,d\varphi$$

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I imagine the projection of your figure on $Oxy$ as upper segment cropped from circle by line $y=x$. So, firstly let's consider volume in Cartesian coordinates $$\int\limits_{0}^{\frac{1}{2}}\int\limits_{x}^{\sqrt{\frac{1}{4}-\left(x-\frac{1}{2}\right)^2}}\int\limits_{0}^{\sqrt{x^2+y^2}}dzdydx$$ So to find borders for cylindrical coordinates we need to solve inequalities

$$\begin{array}{} 0 \leqslant r \cos \phi \leqslant \frac{1}{2} & \\ r \cos \phi \leqslant r \sin \phi \leqslant \sqrt{\frac{1}{4} -\left(r \cos \phi-\frac{1}{2}\right)^2 } & \\ 0 \leqslant z \leqslant r \end{array}$$ From 1-st and 2nd line inequalities we have $0 \leqslant \cos \phi \leqslant \sin \phi$, so we obtain $\phi \in \left[\frac{\pi}{4},\frac{\pi}{2} \right]$ and $0 \leqslant r \leqslant \cos \phi$

$$\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}}\int\limits_{0}^{\cos \phi}\int\limits_{0}^{r}rdzdrd\phi$$

Addition: In case where we consider "sector" of circle i.e. part of circle between lines $y=x$ and $y=0$, then volume in Cartesian coordinates will be

$$\int\limits_{0}^{\frac{1}{2}}\int\limits_{0}^{x}\int\limits_{0}^{\sqrt{x^2+y^2}}dzdydx+\int\limits_{\frac{1}{2}}^{1}\int\limits_{0}^{\sqrt{\frac{1}{4}-\left(x-\frac{1}{2}\right)^2}}\int\limits_{0}^{\sqrt{x^2+y^2}}dzdydx$$

Good news in this case is, that we can calculate this volume again using the founded volume in previous , circle segment, case subtracting it from the volume over the entire semicircle, which is

$$\int\limits_{0}^{1}\int\limits_{0}^{\sqrt{\frac{1}{4}-\left(x-\frac{1}{2}\right)^2}}\int\limits_{0}^{\sqrt{x^2+y^2}}dzdydx=\int\limits_{0}^{\frac{\pi}{2}}\int\limits_{0}^{\cos \phi}\int\limits_{0}^{r}rdzdrd\phi$$ so we obtain $$\int\limits_{0}^{\frac{\pi}{2}}\int\limits_{0}^{\cos \phi}\int\limits_{0}^{r}rdzdrd\phi-\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}}\int\limits_{0}^{\cos \phi}\int\limits_{0}^{r}rdzdrd\phi = \int\limits_{0}^{\frac{\pi}{4}}\int\limits_{0}^{\cos \phi}\int\limits_{0}^{r}rdzdrd\phi$$

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  • $\begingroup$ @Math Lover. I am familiar with this point of view, as Parcly Taxel's answer is 1h before your comment, and I decided to suggest OP possible alternative. My solution is not "between $y$ -axis and the circle", it is up of $y=x$ and down of circle and, as I wrote in first sentence of my answer, is upper segment cropped from circle. Formally $y=0$ also participates (although at one point) in upper segment bounds, so we have 2 possible answers: one Parcly Taxel's, one mine. Can you suggest a formal arguments in your favor? $\endgroup$ – zkutch Nov 26 '20 at 23:27
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    $\begingroup$ I think we should all be grateful for your comment(s), as it makes picture more clear. Meanwhile, I wrote addition, where I linked both approaches. $\endgroup$ – zkutch Nov 27 '20 at 4:31
  • $\begingroup$ That is good addition! $\endgroup$ – Math Lover Nov 27 '20 at 4:37

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