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Surface integrals to find surface area

Suppose we have a plane P defined by the transformation T for minus, 2, is less than, u, is less than, 2 and minus, 1, is less than, v, is less than, 4.

T, left parenthesis, u, comma, v, right parenthesis, equals, left parenthesis, minus, 4, u, plus, 3, v, comma, 2, v, comma, u, plus, 2, v, right parenthesis

What is the surface area of P?

(Choice A)
integral, start subscript, minus, 1, end subscript, start superscript, 4, end superscript, integral, start subscript, minus, 2, end subscript, squared, 2, square root of, 46, end square root, d, u, d, v
(Choice B)
integral, start subscript, minus, 1, end subscript, start superscript, 4, end superscript, integral, start subscript, minus, 2, end subscript, squared, 3, square root of, 21, end square root, d, u, d, v
(Choice C)
integral, start subscript, minus, 1, end subscript, start superscript, 4, end superscript, integral, start subscript, minus, 2, end subscript, squared, 14, d, u, d, v
(Choice D)
integral, start subscript, minus, 1, end subscript, start superscript, 4, end superscript, integral, start subscript, minus, 2, end subscript, squared, 12, d, u, d, v

Stuck?

Multivariable calculus

Course: Multivariable calculus > Unit 4

Surface integrals to find surface area

Multivariable calculus

Course: Multivariable calculus > Unit 4

Problem