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Problem 255

# Consider the transformation $\mathrm{x}-\mathrm{u}+\mathrm{v} \quad \mathrm{y}=\mathrm{v}-\mathrm{u}^{2}$ Let $$D$$ be the set in the $$u-v$$ plane bounded by the lines $\mathrm{u}=0, \mathrm{v}=0$$, and$$\mathrm{u}+\mathrm{v}=2$ Find the area of $$\mathrm{D}^{*}$$, the image of $$\mathrm{D}$$, directly and by a change of variables.

Expert verified
The area of the image of D in the xy-plane is 4 square units, which can be found using the direct method by applying the transformation to the vertices and using the Shoelace formula, or by using a change of variables, computing the Jacobian determinant, and integrating with respect to u and v. In both methods, the resulting area is 4 square units.
See the step by step solution

## Step 1: Parametrize the vertices of D in the uv-plane

Find the points of intersection of the lines bounding D: A: Intersection of $$u = 0$$ and $$v = 0$$: $$(0, 0)$$ B: Intersection of $$u = 0$$ and $$u + v = 2$$: $$(0, 2)$$ C: Intersection of $$v = 0$$ and $$u + v = 2$$: $$(2, 0)$$

## Step 2: Apply the transformation to the vertices

Convert the vertices A, B, and C to their corresponding locations A', B', and C' in the xy-plane using the given transformations: A': $$(0 + 0, 0 - 0^2) = (0, 0)$$ B': $$(0 + 2, 2 - 0^2) = (2, 2)$$ C': $$(2 + 0, 0 - 2^2) = (2, -4)$$

## Step 3: Compute the area of the triangle in the xy-plane (Direct Method)

Use the Shoelace formula to find the area of the triangle formed by A', B', and C' in the xy-plane: Triangle Area\_xyz $$= \frac{1}{2} |(x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A - y_B)|$$ Plug in the coordinates of the transformed vertices and compute the area: Triangle Area\_xyz $$= \frac{1}{2} |(0(2 - (-4)) + 2(-4 - 0) + 2(0 - 2)| = \frac{1}{2} \times 8 = 4$$ So the area of the image of D in the xy-plane is 4 square units.

## Step 4: Compute the area of the triangle in the xy-plane (Using a Change of Variables)

Compute the Jacobian determinant of the given transformation: $$J(u, v) = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \\ \end{vmatrix} = \begin{vmatrix} 1 & 1 \\ -2u & 1 \\ \end{vmatrix} = (1)(1) - (1)(-2u) = 1 + 2u$$ Compute the area of the image of D in the xy-plane using a change of variables and integration: Area\_xyz $$= \iint_D |J(u, v)| \, du \, dv$$ Since D is a right triangle, we can use a double integral to compute its area. Area\_xyz $$= \int_{0}^{2} \int_{0}^{2 - u} (1 + 2u) \, dv \, du = \int_{0}^{2} ((1 + 2u)(2 - u) - (1 + 2u)(0)) \, du$$ Area\_xyz $$= \int_{0}^{2} (2 + 2u - u^2 - 2u^2) \, du = \int_{0}^{2} (2 - 3u^2) \, du$$ Now we can integrate and compute the area: Area\_xyz $$= (2u - u^3)|_0^2 = (4 - 8) = -4$$ The result is negative because D is mapped to the left in the xy-plane. Therefore, we take the absolute value of Area\_xyz to find the actual area: Area\_xyz $$= |-4| = 4$$ So the area of the image of D in the xy-plane using a change of variables is 4 square units, the same as the direct method.

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