**Problem 9: Calculating the Surface Area of a Plane****Objective:**- Set up and evaluate an integral to find the surface area of the portion of the plane given by the equation \( z = mx + ny \).**Region of Interest:**- The plane is evaluated over the rectangle \([0, a] \times [0, b]\).**Constants:**- \( a, b, m, \) and \( n \) are constants.**Instructions:**1. **Integral Setup:** - To find the surface area of the plane over the specified rectangle, use the surface area formula for a plane \( S \): \[ S = \int_0^a \int_0^b \sqrt{1 + \left( \frac{\partial z}{\partial x} \right)^2 + \left( \frac{\partial z}{\partial y} \right)^2} \, dy \, dx \] - For the plane \( z = mx + ny \): \[ \frac{\partial z}{\partial x} = m \quad \text{and} \quad \frac{\partial z}{\partial y} = n \] - Substitute these into the surface area integral: \[ S = \int_0^a \int_0^b \sqrt{1 + m^2 + n^2} \, dy \, dx \]2. **Evaluate the Integral:** - Since the integrand is constant, the integral simplifies: \[ S = \sqrt{1 + m^2 + n^2} \int_0^a \int_0^b 1 \, dy \, dx \] - Calculate the integral: \[ S = \sqrt{1 + m^2 + n^2} \times \bigg[b \times a \bigg] \] - Thus, the surface area \( S \) is: \[ S = ab \sqrt{1 + m^2 + n^2} \]This calculation provides the surface area of the plane over the specified rectangular region on an educational platform.

The given equation of the plane is: z=mx+ny where m and n constants. Let D be the given rectangle. D=0,a×0, bFormula: The surface area A of the function z=f(x, y) over the region D is: A=∫∫Dfx2+fy2+1 dA z=mx+ny Let f(x, y)=mx+ny: Find fx by differentiating f(x,y) partially with respect to x: f(x, y)=mx+nyfx=m Find fy by differentiating f(x,y) partially with respect to y: f(x, y)=mx+nyfy=n A=∫∫Dfx2+fy2+1 dA Substitute the values in the integral: A=∫∫Dm2+n2+1 dA Take the constant out of the integral: A=m2+n2+1∫∫D dAD=0,a×0, b In the region D, x varies from 0 to a and y varies from 0 to b. Hence, ∫∫DdA=∫0a∫0b dy dx The integral for the surface area of the plane z=mx+ny over the rectangle 0,a×0, b is: m2+n2+1∫∫D dA=m2+n2+1∫0a∫0b dy dx∫∫DdA=∫0a∫0b dy dx Evaluate the integral: ∫∫DdA=∫0ay0b dx=∫0ab-0 dx=∫0ab dx=bx0a=ba-0=baA=m2+n2+1∫∫D dA Substitute ∫∫DdA=ab: A=m2+n2+1ab The integral for the surface area of the plane z=mx+ny over the rectangle 0,a×0, b is: m2+n2+1∫0a∫0b dy dx The surface area of the plane z=mx+ny over the rectangle 0,a×0, b is: m2+n2+1ab units