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- I don't understand why the double integral of z dA over region D turns into [(64-4r^2)^(1/2)-(-(64-4r^2)^(1/2)]rdrdtheta, instead of just (64-4r^2)^(1/2) rdrdtheta since R {(r,theta) | 0 <= r <= 2; 0 <= theta <+ 2pi}.Evaluate the triple integral∭E xy dV where E is the solid tetrahedon with vertices (0,0,0),(7,0,0),(0,7,0),(0,0,7).Set up an iterated double integral in polar coordinates equivalentto the area of the surface on the 1st octant that is the portion of theparaboloid x2 + y2 + 2z = 8 below the plane z = 2.
- Say that you need to compute a double integral of the function f(x,y)=xy over the region D bounded by the x-axis, y=x, x2+y2=1, and x2+y2=16. Explain in words and/or show in a picture why this would be (unnecessarily) complicated in Cartesian coordinates. Then, setup and evaluate the integral using polar coordinates.Evaluate the triple integral, 2x2 dV, where T is the solid tetrahedron with verticies (0, 0, 0), (1, 0, 0), (0, 1, 0), and (0, 0, 1)Set up a double integral in polar coordinates to find the volume of the solid bounded above by z=10-4x^2-4y^2 and below by z=4-x^2-y^2. (Do not evaluate).
- [06/09, 11:12 pm] ⚠️: 7-10 Use a double integral in polar coordinates to find the area of the region described. Theregion enclosed by the cardioid r = 1 - cosθEvaluate the iterated integral by converting to polar cooordinates. ∬(e−x2 − y2 ) dy dx limits of outside integral: 0, 4 limits of inside integral: 0, sqrt(16-x2)Calculate the double integral. 6x/1+xy dA, R=[0,6]x[0,1]