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- Q.2/Evaluatethe iterated integral by converting to polar coordinate 2 Vi-yi 5ol K f_y e~ ) dx dyFind an iterated integral in polar coordinates that represents the area of the given region in the polar plane and then evaluate the integral. One loop of the curve r = 4 sin3θ.Consider the triple integral: (ʃ_2^3 ʃ_4^8 ʃ_0^10 sin(xy-z) dxdydz). Use a change of variables to transform the integral to an integral over the cube: {0<a<1, 0<b<1, 0<c<1}.
- Question 1: Identify and sketch the domain of the given function. f(x,y) = sin-1(y-2x)/√1 − x Question 2: Set-up (do not evaluate) the iterated double integral in polar coordinates that gives the area of S. Where we let S be the portion of the paraboloid x2 + y2 = 4 − 2z in the first octant be- tween the planes y = 1 and y = 2.a) Graph the region in the xy-plane to which the integral refers. b) Is it convenient to go to polar coordinates? because? c) Convert to polar coordinates and solve the integral.( a .. )Find the approximations T 8 and M 8 for the integral integral 0 to 1 cos (x 2 ) dx Estimate the errors in the approximations of part (a).
- Use the integration capabilities of a graphing utility to approximate the area of the region bounded by the graph of the polar equation. r = 2/(7 − 6 sin)Graph r = 1 / ( 3 cos θ ) for − π / 2 < θ < π / 2 and r = 1 . Then write an iterated integral in polar coordinates representing the area inside the curve r = 1 and to the right of r = 1 / ( 3 cos θ )Evaluate the triple integral. Note: I'm having trouble understanding why the bounds of dz (0 ≤ z ≤ 3) become (0 ≤ theta ≤ π/2) when switching to polar coordinates.