You may do this project on your own paper. You may work with another person, but you muststate who you worked with and your write-ups must be your own. Please do not copy someoneelse's solution, including solutions found on the internet.One of the application of double integrals is finding the surface area of a region in 3-dimensional space.The book defines surface area and gives the formula to compute surface area.If f and its first partial derivatives are continuous on the closed region Rin the æy-plane, then thearea of the surface S given by z = f (x, y) over Ris defined asSurface area = SrſdS= SRS V1+ [f. (x, y)]² + [fy (x, y)]² dA.Consider the surface area of the plane ax + by + cz + d = 0 completely contained in the firstoctant. (This is assuming a, b, c, and d are all positive real numbers.) Show that the surfacearea of that region on the plane isA(R)Va² + b² + c² where A(R) is the area of the triangle inthe xy-plane in the first octant. Refer to the following picture for guidance.R

Given , If f and its first partial derivative are continuous on the closed region R in the xy-plane, then the area of the surface S is given by z=f(x,y) over R is defined as, Surface area = ∫∫RdS=∫∫R1+fxx,y2+fyx,y2dA. To find the surface area of the plane ax+by+cz+d=0 completely contained in the first octant. To find the surface area: Surface area = ∫∫RdS=∫∫R1+fxx,y2+fyx,y2dA Where R is the projection of the surface in the xy-plane. From the figure it is clear that R is the triangle. ax+by+cz+d=0⇒z=d-ax-byc Thus, fx,y=d-ax-byc fxx,y=-ac, fyx,y=-bc, 1+fxx,y2+fyx,y2=1+-ac2+-bc2=a2+b2+c2c2=1ca2+b2+c2 Surface area = ∫∫RdS =∫∫R1+fxx,y2+fyx,y2dA=∫∫R1ca2+b2+c2dA=1ca2+b2+c2∫∫RdAWe have ∫∫RdA is the area of the region. here R is the triangle.=1ca2+b2+c2×Area of the triangle=ARca2+b2+c2 Hence proved that surface area is ARca2+b2+c2, A(R) is the area of the triangle.