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Find the surface area of the piecewise smooth surface that is the boundary of the region enclosed by the paraboloids z = 9-3x2-3y2 and z-6x2 + 6y2

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User Kamarey
by
8.7k points

2 Answers

5 votes

Final answer:

To find the surface area of the piecewise smooth surface that is the boundary of the region enclosed by the paraboloids z = 9-3x2-3y2 and z-6x2 + 6y2, we need to calculate the surface integral over the region. Specifically, we need to calculate the surface area by converting the double integral into an iterated integral and evaluating it.

Step-by-step explanation:

To find the surface area of the piecewise smooth surface that is the boundary of the region enclosed by the paraboloids z = 9-3x2-3y2 and z-6x2 + 6y2, we need to calculate the surface integral over the region. Specifically, we need to calculate the surface area by converting the double integral into an iterated integral and evaluating it.

First, we need to identify the limits of integration for each variable (x, y, and z) based on the given equations. Then, we set up the double integral, where the integrand is the square root of the partial derivatives of z with respect to x and y squared plus 1, and the limits of integration are determined by the extent of the region.

Finally, we calculate the iterated integral by integrating the square root of the partial derivatives of z with respect to x and y squared plus 1 over the limits of integration. This will give us the surface area of the piecewise smooth surface representing the boundary of the region enclosed by the paraboloids.

answered
User Fenkerbb
by
8.9k points
5 votes

Answer:

z = 3(-6x + 4y + 3)/6y and + 1

Step-by-step explanation:

z = 9 - 3x × 2 - 3y × 2 and z - 6x × 2 + 6y × 2

z = 9 - 3 × 2x - 3 × 2y and z - 6 × 2x + 6 × 2y

z = 9 - 6x - 6y and z - 12x + 12y

z = 3(-6x + 4y + 3)/6y and + 1

answered
User Musaffir Lp
by
8.0k points
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