Evaluate double integral_R 15 x y square root {x^2 + y^2} dA, where R =[0,1] times [0,1].


Evaluate {eq}\displaystyle \iint\limits_R 15 x y \sqrt {x^2 + y^2}\ dA {/eq}, where {eq}R =[0,\ 1] \times [0,\ 1] {/eq}.

Applying Fubini's Theorem on a Rectangle:

Let {eq}f {/eq} be a continuous function on some rectangle (the theorem can be expressed with lesser assumptions on the function f but the function in the problem is continuous so lets solve it like this).

Let the rectangle be {eq}[a,b]\times [c,d]. {/eq}

The Fubini's theorem states that we can evaluate the double integral by simply calculating two Riemann integrals:

{eq}\iint_R f(x,y) \,dx \,dy = \int_a^b \int_c^d f(x,y) \,dy \,dx. {/eq}

We can also switch the order of integration.

When solving the first integral we are integrating over (for example the variable y) and we consider x to be a constant.

Answer and Explanation:

Now lets solve this problem:

{eq}\begin{align*} I&= \int_0^1 \int_0^1 15 x y \sqrt {x^2 + y^2} \,dy \,dx &\text{apply substitution } t= y^2 + x^2 \text{]} \\ &= \int _0 ^1 \frac{15}{2} x \int_{x^2}^{1+x^2} \sqrt{t} \,dt \,dx \\ &=\frac{15}{2} \int _0 ^1 x (\frac{2}{3}t^{\frac{3}{2}}) \Big |_{x^2}^{1+x^2} \,dx \\ &= 5 \int_0 ^1 x(1+x^2)^{\frac{3}{2}} - x^4 \,dx &\text{[evaluate } x^4 \text{ and then apply substitution } t= 1 + x^2 \text{]}\\ &= 5 (\int_1^2 \frac{1}{2} u^{\frac{3}{2}} \,du - \frac{1}{5} ) \\ &= 5 (\frac{1}{2} \frac{2}{5} u^{\frac{5}{2}} \Big |_1^2 - \frac{1}{5} ) \\ &= (2^{\frac{5}{2}} - 1 - 1) \\ &= 2^{\frac{5}{2}} -2 \end{align*} {/eq}

Learn more about this topic:

Double Integrals & Evaluation by Iterated Integrals

from GRE Math: Study Guide & Test Prep

Chapter 15 / Lesson 4

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