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Evaluate the double integral over R of sqrt(x^2 + y^2) dA, where R = (x, y): 1 less than or equal...

Question:

Evaluate {eq}\iint_{R} \sqrt{x^2 + y^2} \, \mathrm{d}A {/eq}, where {eq}R = \left \{ (x, y): 1 \leq x^2 + y^2 \leq 4 \right \} {/eq}.

Definite Integrals:

The definite integral of the double integral form, is resolved in form of limits of integration using the defined region. After getting the integral set up, we need to solve using the polar conversion if applicable. Else we can directly solve using the iteration method.

Answer and Explanation:


Given to us is the definite integral;

{eq}\iint_{R} \sqrt{x^2 + y^2} \, \mathrm{d}A\\ {/eq}

Given region is as follows:

{eq}R = \left \{ (x, y): 1 \leq x^2 + y^2 \leq 4 \right \}\\ {/eq}

So we will form a double integral with the limits of integration as follows:

{eq}\int_{1}^{2}\int_{\sqrt{1-x^2}}^{\sqrt{4-x^2}} \sqrt{x^2 + y^2} dy dx\\ {/eq}

Converting in polar form we have:

{eq}\int _0^{2\pi }\int _1^2\:r^2drd\theta \\ =\int _0^{2\pi }\left(\frac{7}{3}\right)d\theta\\ =\left [ \frac{7}{3}\theta \right ]_0^{2\pi }\\ =\frac{14\pi }{3}\\ {/eq}


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