# Find the total mass of the triangular region with vertices at (-2,0), (0,2) and (2,0). The...

## Question:

Find the total mass of the triangular region with vertices at (-2,0), (0,2) and (2,0). The density is given by {eq}\rho(y)=2+y {/eq} grams per square cm, where {eq}y {/eq} is measured vertically along the y-axis.

## Mass of a Thin Plate using Integration:

Given a thin plate {eq}R{/eq} defined as {eq}R = \left\{\, (x,y) \,|\, c \le y \le d \text{ and } f(y) \le x \le g(y) \,\right\} {/eq} with density {eq}\rho(x,y){/eq}.

The total mass of the thin plate {eq}R{/eq} is given by the double integral:

\begin{align*} &Mass = \iint_R \rho(x,y)\,dA = \int_c^d \int_{f(y)}^{g(y)} \rho(x,y)\,dxdy & \text{[Mass of a Thin Plate with Density } \rho \text{]} \\ \end{align*}

Let the thin plate be defined by the triangular region {eq}R {/eq} with vertices {eq}(-2,0),\, (0,2) \text{ and } (2,0){/eq}.

The graph of the region {eq}R{/eq} is shown below, together with the equations of the lines connecting the vertices written as y in terms of x.

Using the graph, the triangular region can be defined as {eq}R = \left\{\, (x,y) \,|\, 0 \le y \le 2 \text{ and } y - 2 \le x \le -y + 2 \,\right\} {/eq}

Therefore, the total mass of the thin plate with density {eq}\rho(x,y) = (2 + y)\, \frac{g}{cm^2}{/eq} is:

{eq}\begin{align*} Mass &= \int_0^2 \int_{y-2}^{-y+2} (2 + y) \,dxdy & \text{[Integrate with respect to } x \text{]} \\ &= \int_0^2 (2 + y) \left. (x) \right|_{y-2}^{-y+2} \,dy & \text{[Fundamental Theorem of Calculus]} \\ &= \int_0^2 (2 + y) [(-y+2) - (y-2)]\, dy \\ &= \int_0^2 2(2 + y)(2-y)\, dy \\ &= 2 \int_0^2 (4 - y^2)\,dy & \text{[Integrate with respect to } y \text{]} \\ &= 2 \left. \left( 4y - \frac{y^3}{3} \right) \right|_0^2 & \text{[Fundamental Theorem of Calculus]} \\ &= 2 \left[ \left( 4(2) - \frac{(2)^3}{3} \right) - (0) \right] \\ &= 2 \left( \frac{16}{3} \right) \\ Mass &= \boxed{ \frac{32}{3} \approx 10.667 \text{ grams}} \end{align*} {/eq}