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Use the given transformation to evaluate the integral.

∫ ∫R (10x + 10y) dA , where R is the parallelogram with vertices (−2, 8), (2, −8), (4, −6), and (0, 10) ;
X= (u + v)/5, y = (v − 4u)/5

User Penpen
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1 Answer

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The value of the double integral ∫ ∫R (10x + 10y) dA is 224.

To evaluate the double integral using the given transformation, we need

to determine the Jacobian of the transformation and transform the region of integration.

Jacobian of the Transformation

The Jacobian of the transformation is given by:

| J(u, v) | = | ∂x/∂u ∂y/∂u ∂x/∂v ∂y/∂v |

= | 1/5 0 1/5 -4/5 |

= 1/5

Transforming the Region of Integration

The original parallelogram R is defined by the vertices (−2, 8), (2, −8), (4, −6), and (0, 10). Substituting these values into the transformation equations, we get:

(−2, 8) → (0, 6)

(2, −8) → (4, −2)

(4, −6) → (8, −8)

(0, 10) → (0, 10)

The transformed parallelogram R' is defined by the vertices (0, 6), (4, −2), (8, −8), and (0, 10).

Evaluating the Integral

Using the transformed variables u and v, we can now evaluate the double integral:

∫ ∫R (10x + 10y) dA

∫ ∫R' (10(u + v)/5 + 10(v - 4u)/5) |J(u, v)| du dv

= ∫ ∫R' 10u + 10v |1/5| du dv

= ∫ ∫R' 2u + 2v du dv

= (u^2 + 2uv) |_0^8 |_0^{10}

= (64 + 160) - (0 + 0)

= 224

Therefore, the value of the double integral ∫ ∫R (10x + 10y) dA is 224.

User Benjamin Cremer
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