Practicetest-IAP-2010

Practicetest-IAP-2010 - x . Taking F = (6 xy 5 ) i + (1 + x...

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18.02A Practice Test – IAP 2010 Problem 1. a) Write down in xy - coordinates the vector ﬁeld F whose vector at ( x, y ) is obtained by rotating 90 counterclockwise the radially-outward-pointing unit vector at ( x, y ). b) Let F be the ﬁeld in part (a). Let C 1 be the line segment running from (1,1) to (2,2), and C 2 the positively-oriented circle of radius a centered on the origin. Using intuition, give the value of the following (short answer; no calculation required): i) Z C 1 F · d r , ii) I C 2 F · d r , iii) ﬂux of F across C 1 , iv) ﬂux of F across C 2 . Problem 2. Let F = f = grad f , where f ( x, y ) = x 2 + 4 y 2 . a) Evaluate R C F · d r , where C is a curve running from (1,1) to (2,2). b) Find the locus of all points ( x, y ) in the plane such that Z ( x,y ) (1 , 1) F · d r = 0. Problem 3. Let F = y ( ax + y ) i + (3 x 2 + bxy + y 3 ) j , where a, b are constants. a) Prove: if F is conservative, then a = 6 , b = 2. (Use these values in part b). b) Using a systematic method (show work), ﬁnd a function f ( x, y ) such that F = f . Problem 4. Let C be the portion of the parabola y = 1 - x 2 lying over the x - axis, oriented in the direction of decreasing
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Unformatted text preview: x . Taking F = (6 xy 5 ) i + (1 + x 2 y-y 6 ) j , a) set up an integral in x alone that represents the ux of F over C . (Give integrand and limits, but do not evaluate); b) calculate the ux of F over C by using Greens Theorem in the normal form. (Note that C is not closed). Problem 5. Show that the value of I C ( y 2-2 y ) dx + 2 xy dy around a positively oriented circle C depends only on the size of the circle, and not on its position. Problem 6. Consider the integral Z Z R ( x + y ) 4 (3 x-y ) 4 dxdy , where R is the triangle with vertices at x =-1 and x = 3 on the x-axis, and y = 3 on the y-axis. Let u = x + y and v = 3 x-y . Express the double integral in uv-coordinates; use as the order of integration dv du . Problem 7. Calculate the volume in the upper half plane contained between the cone, z = ( x 2 + y 2 ) 1 / 2 and the sphere x 2 + y 2 + z 2 = 2....
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This note was uploaded on 05/14/2010 for the course 18.02A 18.02A taught by Professor Johnbush during the Winter '10 term at MIT.

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