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Chapter 12: Problem 9

Find the area of the surface. The portion of the plane \(z=24-3 x-2 y\) in the first octant

Short Answer

Expert verified
The area of the surface of the plane in the first octant is \(48 \sqrt{14}\).

Step by step solution

01

Determine the partial derivatives

We first need to determine the partial derivatives of the function \(z\) with respect to \(x\) and \(y\): \( z = 24 - 3x - 2y \) So, \( \frac{dz}{dx} = -3 \) and \( \frac{dz}{dy} = -2 \)
02

Find the differential element of the surface

Next, we proceed to find the differential element of the surface \(ds\). From the surface element \( ds = \sqrt{1 + (\frac{dz}{dx})^2 + (\frac{dz}{dy})^2} \,dxdy \) substituting the values we get, \( ds = \sqrt{1 + ( -3 )^2 + ( -2 )^2} \,dxdy = \sqrt{1 + 9 + 4} \,dxdy = \sqrt{14} \,dxdy \)
03

Find the limits for the integral

The limits of the integral are determined by the boundaries of the first octant, that is where \(x, y, z \geq 0\). The plane intersects the x-axis where \(y=z=0\), so \(x = 8\). Similarly, when \(x=z=0\), we get \(y=12\). Therefore, limits for \(x\) are \([0, 8]\) and for \(y\) are \([0, 12-y]\).
04

Compute the integral

Now we compute the double integral to find the surface area: \(A = \int_0^{8} \int_0^{12-x} \sqrt{14} \,dy dx = \sqrt{14} \int_0^{8} (12 - x) dx \). Then do the integral computation, the result is \(A = \sqrt{14} \left. \left[ 12x - \frac{x^2}{2} \right] \right|_0^8 = 48 \sqrt{14}\).

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