How do we find the area of the surface of the part of the plane 2xy + 3z = 1 that lies inside the cylinder x^2 + y^2 = 3?

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To find the area of the surface of the part of the plane defined by the equation 2xy + 3z = 1 that lies within the cylinder described by x^2 + y^2 = 3, we can follow these steps:

  1. Rearranging the Plane Equation: First, we need to express the plane equation in terms of z, thus making it easier to work with:

    2. z = (1 – 2xy) / 3

  2. Finding the Partial Derivatives: Next, we’ll calculate the partial derivatives of z with respect to x and y:

    3. ∂z/∂x = -2y / 3

    ∂z/∂y = -2x / 3

  3. Calculating the Area Element: The formula for the area element dS on the surface z = f(x,y) is given by:

    4. dS = sqrt(1 + (∂z/∂x)2 + (∂z/∂y)2) dx dy

    Substituting in our partial derivatives:

    dS = sqrt(1 + (-2y/3)2 + (-2x/3)2) dx dy

    dS = sqrt(1 + (4y2/9) + (4x2/9)) dx dy

  4. Setting Up the Integral: Now we need to set up a double integral over the region defined by the cylinder x2 + y2 <= 3:

    5. Area = ∬R dS

  5. Changing to Polar Coordinates: To compute the integral more easily, we change to polar coordinates where:

    6. x = r cos(θ), y = r sin(θ)

    Here, R is defined as 0 <= r <= sqrt(3) and 0 < θ < 2π.

  6. Calculating the Area in Polar Coordinates: After substituting x and y into the area element, the integral becomes:

    7. Area = ∬ (sqrt(1 + (4 (r sin(θ))2/9) + (4 (r cos(θ))2/9)) r dr dθ

  7. Integrating: We need to evaluate the integral to find the total area:

    8. Area = ∫2π0 ∫sqrt(3)0 (r sqrt(1 + (4r2 (sin2(θ) + cos2(θ)) / 9)) dr dθ

Once the integral is computed, it will give you the area of the surface of the plane that lies within the specified cylinder.

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