Question

1. The region W is the cone shown below. The angle at the vertex is 2π/3, and the top is flat and at a height of 3/√3.

Write the limits of integration for ∫WdV in the following coordinates (do not reduce the domain of integration by taking advantage of symmetry):

2. Match each vector field with its graph.

I know we are only supposed to post 1 per question however for this one I have 1 part correct, I just need some help with the rest. Please if you have the time help with question 2. Thank you for your time and knowledge.

I am sorry I dont Know what "In sub" means to the person who made a comment" Let me know what needs clarification.

(1 point) The region W is the cone shown below. The angle at the vertex is 2/3, and the top is flat and at a height of 3/V3. Write the limits of integration for Sw dV in the following coordinates (do not reduce the domain of integration by taking advantage of symmetry): (a) Cartesian: With a - ,b c ,d e and S Volume = LOSSO d (b) Cylindrical: With a = b= C= .d c and Volume = Session Set d d d (c) Spherical: With a b= d= e and f Volume = Sou See Seel d d d

Answer 1

1)a)Angle at vertex = $\frac{2\pi}{3}$ and top is flat and height = $\frac{3}{\sqrt{3}}$

The cone has an angle of $\frac{2\pi}{3}$ at its vertex and the angle between the cone and the zaxis is half of this

3

$0\leq \phi \leq \frac{\pi}{3}$

cone height = $\frac{3}{\sqrt{3}}$

3 0<:< /3

So

$R= \frac{3}{\sqrt{3}}tan(60)=3\\ \sqrt{x^2+y^2}=3\ x^2+y^2=9\\ y=\pm \sqrt{9-x^2}\\ x\epsilon [-3,3]$

$\large V=\int_{-3}^{3}\int_{- \sqrt{9-x^2}}^{ \sqrt{9-x^2}}\int_{\frac{\sqrt{x^2+y^2}}{\sqrt{3}}}^{\frac{3}{\sqrt{3}}}1dzdydx$

b) In terms of cylindrical coordinates, we need r

tano psino pcoso (x² + y² 7T tan - = 3 2 1 x² + y² V3 z? + y2 = V32 T= V3:

$V=\int_{0}^{2\pi}\int_{0}^{\frac{3}{\sqrt{3}}}\int_{0}^{z\sqrt{3}}rdrdzd\theta$

$z\sqrt{3}=r\\ z=\frac{r}{\sqrt{3}}\\ z=\frac{3}{\sqrt{3}}\\ r=3$

$\large V=\int_{0}^{2\pi}\int_{0}^{3}\int_{\frac{r}{\sqrt{3}}}^{\frac{3}{\sqrt{3}}}rdzdrd\theta$

c) spherical coordinates

the cone has equation $\phi= \frac{\pi}{3}$ and the plane $z=\frac{3}{\sqrt{3}}$

$\rho cos\phi =\frac{3}{\sqrt{3}}\\ \rho =\frac{3}{\sqrt{3}cos\phi }\\$

The volume in spherical coordinates is:

$\large V=\int_{0}^{2\pi}\int_{0}^{\frac{\pi}{3}}\int_{0}^{\frac{3}{\sqrt{3}cos\phi }}\rho ^2sin\phi d\rho d\phi d\theta$​​​​​​​

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