# Suppose that f(x,y) = 40,000 y e^{-x^2-y^2} models the population density (population per square...

## Question:

Suppose that {eq}f(x,y) = 40,000 y e^{-x^2-y^2} {/eq} models the population density (population per square mile) of a species of small animals, with x and y measured in miles. Estimate the population in the triangular-shaped habitat with vertices (1,1), (2,1), and (1,0). Round your answer to the nearest whole number.

## Double Integrals

The integral of {eq}\displaystyle f(x,y) {/eq} over a region described in Cartesian coordinates {eq}\displaystyle \mathcal{R}=\{(x,y)|\, a\leq y\leq b, u(y)\leq x\leq v(y)\} {/eq} is

{eq}\displaystyle \iint_{\mathcal{R}}f(x,y)\ dA {/eq} and evaluated iteratively as {eq}\displaystyle \int_a^b\int_{u(y)}^{v(y)}f(x,y) \ dxdy. {/eq}

If the integrand function is not easy to integrate with respect to {eq}\displaystyle x, {/eq} then we need to change the order of integration, by describing the region as:

{eq}\displaystyle \displaystyle \mathcal{R}=\{(x,y)|\, c\leq x\leq d, w(x)\leq y\leq h(x)\} {/eq}

and the integral is {eq}\displaystyle \int_c^d\int_{w(x)}^{h(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 the population density function {eq}\displaystyle f(x,y) = 40,000 y e^{-x^2-y^2} {/eq}

we will evaluate the following double integral {eq}\displaystyle \iint_{\mathcal{D}} f(x,y)\ dA \left[ \frac{\text{ population}}{mi^2}\cdot mi^2=\text{ population}\right]. {/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 connecting {eq}\displaystyle \mathbf{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 40,000 y e^{-x^2-y^2} \ dydx\\ &=-\int_1^2 20,000 e^{-x^2-y^2}\bigg\vert_{y=x-1}^{y=1} \ dx, &\left[ \text{using the substitution } u=-x^2-y^2, du=-2y\ dy \right]\\ &=-\int_1^2 20,000 \left[e^{-x^2-1}-e^{-x^2-(x-1)^2}\right]\ dx\\ &=20,000\int_1^2 \left(e^{-x^2-(x-1)^2}-e^{-x^2-1}\right)\ dx\\ &\boxed{\approx 1,396 \text{ species }}, &\left[\text{using numerical integration, limit of a Riemann sum}\right]. \end{align} {/eq}