# Evaluate the following double integral \iint_{R} 7x^{3}y dA where R is the rectangle with...

## Question:

Evaluate the following double integral

{eq}\displaystyle\; \iint_{R} 7x^{3}y \,dA {/eq}

where {eq}R {/eq} is the rectangle with vertices {eq}(0,0),\; (2,0),\; (0,6), {/eq} and {eq}(2,6) {/eq}.

## Evaluating Double Integral:

The given function is used for evaluating the double integration over the limits. We have to integrate the function twice with respect to the variables. The rectangular vertices are given. From that, we can get the limits.

Given condition is {eq}\displaystyle\; \iint_{R} 7x^{3}y \,dA {/eq} and the rectangle vertices are {eq}\displaystyle (0,0),\; (2,0),\; (0,6), {/eq} and {eq}\displaystyle (2,6) {/eq}.

So the limits are {eq}\displaystyle 0 \leq x \leq 2 {/eq} and {eq}\displaystyle 0 \leq y \leq 6 {/eq}.

Evaluating the double integral:

{eq}\begin{align*} \displaystyle \text{I} &=\iint_{R} \left( 7x^{3}y \right) \,dA \\ \displaystyle &=\int_{0}^{6}\int_{0}^{2} \left( 7x^{3}y \right) \ dxdy \\ \displaystyle &=\int_{0}^{6} \left[ \frac{7yx^4}{4} \right]_{0}^{2} \ dy \\ \displaystyle &=\int_{0}^{6} \left[ \left( \frac{7y(2)^4}{4} \right)-\left( \frac{7y(0)^4}{4} \right) \right] \ dy \\ \displaystyle &=\int_{0}^{6} (28y) \ dy \\ \displaystyle &=\left[ 14y^2 \right]_{0}^{6} \\ \displaystyle &=(14(6)^2)-(14(0)^2) \\ \displaystyle \text{I} &=504 \end{align*} {/eq}

The answer for the double integral is {eq}\ \displaystyle \boxed{\mathbf{\color{blue}{ I =504 }}} {/eq}. 