# Suppose that f(x,y) = 50,000y e^{-x^2 - y^2} models the population density (population per...

## Question:

Suppose that {eq}f(x,y) = 50,000y\; e^{-x^2 - y^2} {/eq} models the population density (population per square mile) of a species of small animals, with {eq}x {/eq} and {eq}y {/eq} measured in miles.

Estimate the population in the triangular shaped habitat with vertices {eq}(1,1), \;(2,1) {/eq} and {eq}(1,0) {/eq}.

Round the answer to the nearest whole number.

## Double Integrals

To find the population of a species over a given region {eq}\displaystyle \mathcal{D}, {/eq} knowing the density function, {eq}\displaystyle f(x,y) {/eq} as number of species per square mile,

we will integrate the density function over the region, so we will evaluate a double integral whose value will be the population of the species.

To evaluate a double integral, {eq}\displaystyle \iint_{\mathcal{R}}f(x,y)\ dA {/eq} we need to describe the region of integration in Cartesian or polar coordinates.

In Cartesian coordinates, we can describe the region as a {eq}\displaystyle x- {/eq} simple or {eq}\displaystyle y- {/eq} simpel region.

If the region is described as a{eq}\displaystyle y- {/eq} simple region, {eq}\displaystyle \mathcal{R}=\{(x,y)|\, a\leq x\leq b, u(x)\leq y\leq v(x)\} {/eq} then the integral is evaluated iteratively as {eq}\displaystyle \int_a^b\int_{u(x)}^{v(x)}f(x,y) \ dydx. {/eq}

To find the population in the triangular-shaped habitat {eq}\displaystyle \mathcal{D} {/eq} with vertices {eq}\displaystyle (1,1), (2,1), \text{ and } (1,0), {/eq} knowing that the population density function is {eq}\displaystyle f(x,y) = 50,000 y e^{-x^2-y^2}, {/eq}

we will evaluate the following double integral {eq}\displaystyle \iint_{\mathcal{D}} f(x,y)\ dA. {/eq}

For this, we need to describe the region of integration in Cartesian coordinates.

The triangular-shaped region is bounded above by the line connecting {eq}\displaystyle (1,1) \text{ and } (2,1): y=1, {/eq}

below by the line connecting {eq}\displaystyle (1,0) \text{ and } (2,1): y=x-1, {/eq}

between {eq}\displaystyle x=1 \text{ and }x=2, {/eq}

so the region is described in Cartesian coordinates as {eq}\displaystyle \mathcal{D}=\{(x,y)| \ 1\leq x\leq 2, x-1\leq y\leq 1\}. {/eq}

Therefore the double integral is calculated as below,

{eq}\displaystyle \begin{align} \iint_{\mathcal{D}} f(x,y)\ dA&=\int_1^2 \int_{x-1}^1 50,000 y e^{-x^2-y^2} \ dydx\\ &=-\int_1^2 25,000 e^{-x^2-y^2}\bigg\vert_{y=x-1}^{y=1} \ dx, &\left[ \text{using the substitution } u(y)=-x^2-y^2, du=-2y\ dy \right]\\ &=-\int_1^2 25,000 \left[e^{-x^2-1}-e^{-x^2-(x-1)^2}\right]\ dx\\ &=25,000\int_1^2 \left(e^{-x^2-(x-1)^2}-e^{-x^2-1}\right)\ dx\\ &\boxed{\approx 1,745 \text{ species }}, &\left[\text{using numerical integration, limit of a Riemann sum}\right]. \end{align} {/eq}