Let I = , S, xdxdy. Our aim is to convert I into an equivalent polar integral. Then the limits of integration of 0 in the order drd0 are: T/45OST/2 OSOST/4 O None of these OSOST/2
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- Evaluate the following integrals by converting to polar coordinates: (a) / VI=x² (1+ x² + y2)2dydx 2 (b) L eV²+y² dxdy /2-x² (c) /L (x+ y) dydxEvaluate the iterated integral by converting to polar coordinates 16-у2 (x² + y²) dx dyEvaluate the iterated integral -6x-x2 XY dy dx, (x² + y?}3/2 -3 Jo using polar coordinates.
- O (-3,-1) VE Consider the double integral, SN * dxdy. By converting to polar coordinates, the limits of integration ofr are: None of these O Osrs2 O Osrs3 O Osrsv2 V9-x2 Cauridartha double integral dydx. By conyerting to polar coordinates,Evaluate the integral by converting to polar coordinates. /8-y² 2 [² [²² 1 2 √1 + x² + y² dx dyLet C be the circumference x2 + y2 = 4x. Consider the following line integral:(img15) The double integral that results from applying Green's Theorem, written in polar coordinates is: Select one:(img16)
- zain IQ I. Q1.jpg Q1: 3 1 SS e dydx a) By using the change of order evaluate the integral Vx /3 3 19-x 2 b) Use the polar coordinate to find the integral | sin(x +y²) dydx -3Calculate the area bounded by r = 2 cos 0 and the rays 0 = 0 and 0 =7 as an integral in polar coordinates. 3.Sketch the solid of integration of D only