# Suppose W is the solid region above the cone z = \sqrt{x^2+ y^2} and below the plane z = 3 ,...

## Question:

Suppose W is the solid region above the cone {eq}z = \sqrt{x^2+ y^2} {/eq} and below the plane {eq}z = 3 {/eq}, where {eq}y \geq 0 {/eq}. The density over W is given by the function {eq}\delta(x, y, z) = z {/eq}

Set up and evaluate a triple integral that gives the mass of W using Cartesian coordinates

## Integration:

A mechanical quantity that represents the merge of the differential of the function to form the original function is known as integration. It used in physics applications to solve the system equation.

Given Data:

• {eq}z = \sqrt {{x^2} + {y^2}} {/eq}
• {eq}z = 3 {/eq}
• The density is: {eq}\delta \left( {x,y,z} \right) = z {/eq}

{eq}{R^2} = {x^2} + {y^2} {/eq}

Substitute the value and solve the above expression

{eq}\begin{align*} R &= \sqrt {{x^2} + {y^2}} \\ &= \sqrt z \\ &= \sqrt 3 \end{align*} {/eq}

The expression small volume in polar coordinates is

{eq}dV = zdzdRd\theta {/eq}

The expression for mass of solid is

{eq}dm = \delta \left( {x,y,z} \right)zdzdRd\theta \cdots\cdots\rm{(I)} {/eq}

The limits of variable is

{eq}\begin{align*} 0 &\le z \le 3\\ 0 &\le R \le \sqrt 3 \\ 0 &\le \theta \le 2\pi \end{align*} {/eq}

Substitute the value and integrate the expression (I) with respective limits

{eq}\begin{align*} \int {dm} &= \int_0^{2\pi } {\int_0^{\sqrt 3 } {\int_0^3 {{z^2}dzdRd\theta } } } \\ m &= \int_0^{2\pi } {\int_0^{\sqrt 3 } {\left[ {\dfrac{{{z^3}}}{3}} \right]_0^3dRd\theta } } \\ &= \int_0^{2\pi } {\int_0^{\sqrt 3 } {\left[ {\dfrac{{{{\left( 3 \right)}^3}}}{3} - 0} \right]dRd\theta } } \\ &= \int_0^{2\pi } {d\theta } \int_0^{\sqrt 3 } {9dR} \\ &= 9\left[ \theta \right]_0^{2\pi }\left[ R \right]_0^{\sqrt 3 }\\ &= 9\left[ {2\pi - 0} \right]\left[ {\sqrt 3 - 0} \right]\\ &= 18\sqrt 3 \pi \end{align*} {/eq}

Thus the mass of solid region is {eq}18\sqrt 3 \pi {/eq}