# 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.

## Answer and Explanation:

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}.

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#### Learn more about this topic:

from AP Calculus AB & BC: Help and Review

Chapter 12 / Lesson 15