The area enclosed by the curve |x + y| + |x - | vedclass

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(A) The given equation is $|x + y| + |x - y| = 11$.Let $u = x + y$ and $v = x - y$. The equation becomes $|u| + |v| = 11$.This represents a square in

Vedclass · Accepted Answer

$121$

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(A) The given equation is $|x + y| + |x - y| = 11$.Let $u = x + y$ and $v = x - y$. The equation becomes $|u| + |v| = 11$.This represents a square in the $uv$-plane with vertices at $(11, 0), (0, 11), (-11, 0), (0, -11)$.The area of this square in the $uv$-plane is $2 	imes (11)^2 = 242$.The transformation from $(x, y)$ to $(u, v)$ is given by $u = x + y$ and $v = x - y$.The Jacobian of this transformation is $J = \begin{vmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{vmatrix} = \begin{vmatrix} 1 & 1 \ 1 & -1 \end{vmatrix} = -1 - 1 = -2$.The area in the $xy$-plane is given by $	ext{Area}_{xy} = \frac{	ext{Area}_{uv}}{|J|} = \frac{242}{|-2|} = \frac{242}{2} = 121$.Alternatively,the equation $|x + y| + |x - y| = a$ represents a square with vertices at $(\pm a/2, \pm a/2)$. Here $a = 11$,so vertices are $(\pm 5.5, \pm 5.5)$.The side length of this square is the distance between $(5.5, 5.5)$ and $(-5.5, 5.5)$,which is $11$.Thus,the area is $11^2 = 121$.

The area enclosed by the curve $|x + y| + |x - y| = 11$ is

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