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Evaluate the integral integral integral_R squareroot 2x(y - 2x) dA, where the region R = {(x, y);...

Question:

Evaluate the integral {eq}\displaystyle \int \int_R \sqrt {2x(y - 2x)} \ dA, {/eq} where the region {eq}R = \{ (x, y); 0 \leq x \leq 2, 2x \leq y \leq 2x + 1 \}. {/eq}

Use the transformation {eq}T: x = 2u, y = 4u + v {/eq}

Change of Variables in Integration:

Given the integral of {eq}F(x,y) {/eq} over the region {eq}R {/eq} in the xy-plane expressed as

$$\iint \limits_{R(x,y)} F(x,y)\,dA $$

Let the variables x and y be expressed in terms of u and v using the transformation equations:

$$x = g(u,v) \\ y = h(u,v) $$

Therefore, the double integral in the xy-plane written above can be expressed as a double integral in the uv-plane as:

$$\iint \limits_{R(x,y)} F(x,y)\,dA = \iint \limits_{R(u,v)} F(u,v)\,dA \\ dA = dx\,dy = J\,du\,dv $$

where {eq}J {/eq} is the Jacobian of the transformation and is given by the partial derivatives of the transformation equations:

$$J = det \begin{vmatrix} \frac{dx}{du} & \frac{dx}{dv} \\ \frac{dy}{du} & \frac{dy}{dv} \end{vmatrix} = \frac{dx}{du} \cdot \frac{dy}{dv} - \frac{dx}{dv} \cdot \frac{dy}{du} \\ $$

Answer and Explanation:

Given the double integral {eq}\iint_R \sqrt{ 2x ( y - 2x) }\,dA {/eq} where the region {eq}R = \left\{\, (x,y) \,|\, 0 \le x \le 2 \text{ and } 2x \le y \le 2x + 1 \,\right\} {/eq} .

Also given is the transformation {eq}T: x = 2u \text{ and } y = 4u + v {/eq}.


Limits of Integration

Solve for the new limits of integration using the the given limits of {eq}R {/eq}:

{eq}\begin{align*} &0 \le x \le 2 & \text{[Substitute } x = 2u ] \\ &\Rightarrow 0 \le 2u \le 2 & \text{[Multiplication Property of Inequality]} \\ &\Rightarrow \boxed{ 0 \le u \le 1 } & \text{[Limits of } u ] \\ \\ &2x \le y \le 2x + 1 & \text{[Substitute } x = 2u \text{ and } y = 4u + v ] \\ &\Rightarrow 2(2u) \le 4u + v \le 2(2u) + 1 \\ &\Rightarrow 4u \le 4u + v \le 4u + 1 & \text{[Addition Property of Inequality]} \\ &\Rightarrow \boxed{ 0 \le v \le 1 } & \text{[Limits of } v ] \\ \end{align*} {/eq}


Integrand

The integrand when expressed in the uv-plane is:

{eq}\begin{align*} \sqrt{2x(y - 2x)} &= \sqrt{ 2(2u) [ (4u + v) - 2(2u) ] } \\ &= \sqrt{ 4u ( 4u + v - 4u ) } \\ &= \sqrt{ 4uv } \\ &= 2 \sqrt{ uv } & \text{[The integrand in the } uv \text{-plane]} \end{align*} {/eq}


Jacobian of the Transformation

Solve for the partial derivatives of the transformation equations to solve for the Jacobian of the transformation:

{eq}\begin{align*} \frac{dx}{du} &= \frac{d}{du} ( 2u ) \\ &= 2 \\ \frac{dx}{dv} &= \frac{d}{dv} ( 2u ) \\ &= 0 \\ \frac{dy}{du} &= \frac{d}{du} ( 4u + v ) \\ &= 4 \\ \frac{dy}{dv} &= \frac{d}{dv} ( 4u + v ) \\ &= 1 \\ J &= \frac{dx}{du} \cdot \frac{dy}{dv} - \frac{dx}{dv} \cdot \frac{dy}{du} \\ &= 2 \cdot 1 - 0 \cdot 4 \\ &= 2 & \text{[Jacobian of the Transformation]} \end{align*} {/eq}


Integral

Therefore, the given double integral in the xy-plane can be evaluated as the double integral in the uv-plane:

{eq}\begin{align*} \iint_R \sqrt{ 2x ( y - 2x) }\,dA &= \int_0^1 \int_0^1 2 \sqrt{uv} (2\,dudv) \\ &= \int_0^1 \int_0^1 4 u^\frac{1}{2} v^\frac{1}{2} \,dudv & \left[ \text{For all constant } k ,\, \int kf(x)dx = k \int f(x)dx \right] \\ &= 4 \int_0^1 \int_0^1 u^\frac{1}{2} v^\frac{1}{2} \,dudv & \text{[Integrate with respect to } u \text{]} \\ &= 4 \int_0^1 v^\frac{1}{2} \left. \left( \frac{2}{3} u^\frac{3}{2} \right) \right|_0^1 \,dv & \text{[Fundamental Theorem of Calculus]} \\ &= 4 \int_0^1 \frac{2}{3} v^\frac{1}{2} \left( 1^\frac{3}{2} - 0^\frac{3}{2} \right) \,dv \\ &= \frac{8}{3} \int_0^1 v^\frac{1}{2} \,dv & \text{[Integrate with respect to } v \text{]} \\ &= \frac{8}{3} \left. \left( \frac{2}{3} v^\frac{3}{2} \right) \right|_0^1 & \text{[Fundamental Theorem of Calculus]} \\ &= \frac{16}{9} \left( 1^\frac{3}{2} - 0^\frac{3}{2} \right) \\ \iint_R \sqrt{ 2x ( y - 2x) }\,dA &= \boxed{ \frac{16}{9} \approx 1.778 } \end{align*} {/eq}


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How to Solve Integrals Using Substitution

from Math 104: Calculus

Chapter 13 / Lesson 5
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