Chapter 13: Q 25. (page 1066)

Give the rectangular and spherical coordinates for the point with cylindrical coordinates:
$(\sqrt{48}, \frac{π}{3}, 4)$

Short Answer

Expert verified

Rectangular coordinates are $(2 \sqrt{3}, 6, 4)$ and spherical coordinates are $(8, \frac{π}{3}, \frac{π}{3})$

Step by step solution

Given Information

The given expression is $(r, θ, ϕ) = (\sqrt{48}, \frac{π}{3}, 4)$

Finding the rectangular coordinates

As $r = \sqrt{48}, θ = \frac{π}{3}, ϕ = 4$

To find out rectangular coordinates, we will use

$x = r \cos θ, y = r \sin θ, z = z$

Using the values

$x = \sqrt{48} \cos \frac{π}{3}, y = \sqrt{48} \sin \frac{π}{3}, z = 4$

$x = 2 \sqrt{3}, y = 6, z = 4$

$(x, y, z) = (2 \sqrt{3}, 6, 4)$

Finding the spherical coordinates

To convert rectangular to spherical coordinates

$ρ = {\sqrt{r^{2} + z^{2}}, θ = θ, ϕ = \tan}^{- 1} \frac{r}{z}$

Putting the values

$ρ = {\sqrt{{(\sqrt{48})}^{2} + 4^{2}}, θ = \frac{π}{3}, ϕ = \tan}^{- 1} \frac{\sqrt{48}}{4}$

$ρ = {\sqrt{48 + 16}, θ = \frac{π}{3}, ϕ = \tan}^{- 1} \sqrt{3}$

$\Rightarrow ρ = \sqrt{64}, θ = \frac{π}{3}, ϕ = \frac{π}{3}$

$\Rightarrow ρ = 8, θ = \frac{π}{3}, ϕ = \frac{π}{3}$

$\Rightarrow (ρ, θ, ϕ) = (8, \frac{π}{3}, \frac{π}{3})$

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Give the rectangular and spherical coordinates for the point with cylindrical coordinates:
$(\sqrt{48}, \frac{π}{3}, 4)$

Short Answer

Step by step solution

Given Information

Finding the rectangular coordinates

Finding the spherical coordinates

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Give the rectangular and spherical coordinates for the point with cylindrical coordinates:
$(\sqrt{48}, \frac{π}{3}, 4)$