Violet Woodward

2022-07-17

Volume Integration of Bounded Region

I'm trying to integrate this volume in spherical and cylindrical coordinates, but having difficulty finding my bounds of integration;

I'm given the region in the first octant bounded by $z=$ $\sqrt{{x}^{2}+{y}^{2}}$, $z=$ $\sqrt{1-{x}^{2}-{y}^{2}}$, $y=x$ and $y=$ $\sqrt{3}x$ and I need to evaluate ${\iiint}_{V}dV$.

When proceeding to integrate with spherical and cylindrical coordinates I am not getting the right bounds such that both methods equate to the same volume? I am definitely missing something. Any and all advice would be much appreciated!

I'm trying to integrate this volume in spherical and cylindrical coordinates, but having difficulty finding my bounds of integration;

I'm given the region in the first octant bounded by $z=$ $\sqrt{{x}^{2}+{y}^{2}}$, $z=$ $\sqrt{1-{x}^{2}-{y}^{2}}$, $y=x$ and $y=$ $\sqrt{3}x$ and I need to evaluate ${\iiint}_{V}dV$.

When proceeding to integrate with spherical and cylindrical coordinates I am not getting the right bounds such that both methods equate to the same volume? I am definitely missing something. Any and all advice would be much appreciated!

yermarvg

Beginner2022-07-18Added 19 answers

Step 1

This region seems better defined using spherical coordinates than cylindrical. We are given that the region is between two vertical planes $y=x$ and $y=\sqrt{3}x$, and it is between the sphere ${x}^{2}+{y}^{2}+{z}^{2}=1$ and the upper half of the cone ${x}^{2}+{y}^{2}={z}^{2}$.

Step 2

From this, we can set the bounds to be:

$\frac{\pi}{3}\le \theta \le \frac{\pi}{4}$

from the region of angles between the two lines (arctan of root(3) is pi/3) $0\le \varphi \le \frac{\pi}{4}$ from the intersection of the cone and sphere $0\le r\le 1$ from the radius of sphere

This region seems better defined using spherical coordinates than cylindrical. We are given that the region is between two vertical planes $y=x$ and $y=\sqrt{3}x$, and it is between the sphere ${x}^{2}+{y}^{2}+{z}^{2}=1$ and the upper half of the cone ${x}^{2}+{y}^{2}={z}^{2}$.

Step 2

From this, we can set the bounds to be:

$\frac{\pi}{3}\le \theta \le \frac{\pi}{4}$

from the region of angles between the two lines (arctan of root(3) is pi/3) $0\le \varphi \le \frac{\pi}{4}$ from the intersection of the cone and sphere $0\le r\le 1$ from the radius of sphere

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