Final 2003

# Final 2003 - MATH 52 FINAL EXAM SOLUTIONS(AUTUMN 2003 1...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: MATH 52 FINAL EXAM SOLUTIONS (AUTUMN 2003) 1. Evaluate the integral by reversing the order of integration integraldisplay 1 integraldisplay 3 3 y e x 2 dxdy. Solution. integraldisplay 1 integraldisplay 3 3 y e x 2 dxdy = integraldisplay 3 integraldisplay x/ 3 e x 2 dxdy = integraldisplay 3 bracketleftBig e x 2 y bracketrightBig y = x/ 3 y =0 dx = integraldisplay 3 parenleftBig x 3 parenrightBig e x 2 dx = 1 6 e x 2 vextendsingle vextendsingle vextendsingle vextendsingle 3 = e 9- 1 6 2. Rewrite the following integral in rectangular coordinates. integraldisplay π/ 4 integraldisplay 2 cos θ +sin θ r 2 cos θ dr dθ Do not evaluate the integral. Solution. The equation r = 2 / (cos θ + sin θ ) becomes x + y = 2 in rectangular co- ordinates. Since r drdθ becomes dA , the integrand r cos θ becomes x in rectangular coordinates, so the integral equals integraldisplay 1 integraldisplay 2- y y xdxdy or integraldisplay 1 integraldisplay x x dydx + integraldisplay 2 1 integraldisplay 2- x xdydx 3. (a) Consider the change of variables defined by u = y x , v = xy. Find the Jacobian ∂ ( x, y ) ∂ ( u, v ) in terms of u and v . Solution. First we find the Jacobian ∂ ( u,v ) ∂ ( x,y ) . It is equal to the determinant bracketleftbigg- y/x 2 y 1 /x x bracketrightbigg =- 2 y/x =- 2 u Hence the Jacobian ∂ ( x, y ) ∂ ( u, v ) =- 1 2 u . 1 (b) Use your answer to part (a) to find the area of the region in the first quadrant bounded by the lines y = x and y = 2 x and the curves xy = 2 and xy = 3. Solution. In the uv-coordinates the region becomes 1 ≤ u ≤ 2 and 2 ≤ v ≤ 3. Hence its area is: integraldisplay 2 u =1 integraldisplay 3 v =2 1 2 u dv du = integraldisplay 2 u =1 1 2 u du = bracketleftbigg 1 2 ln u bracketrightbigg 2 u =1 = 1 2 ln2 4. Find the area of the part of hyperbolic paraboloid z = y 2- x 2 that lies between the cylinders x 2 + y 2 = 1 and x 2 + y 2 = 4. Solution. z = f ( x, y ) = y 2- x 2 with 1 ≤ x 2 + y 2 ≤ 4. Then A ( S ) = integraldisplayintegraldisplay D radicalbig 1 + 4 x 2 + 4 y 2 dA = integraldisplay 2 π integraldisplay 2 1 √ 1 + 4 r 2 r drdθ = integraldisplay 2 π dθ integraldisplay 2 1 √ 1 + 4 r 2 r dr = 2 π bracketleftbigg 1 12 (1 + 4 r 2 ) 3 / 2 bracketrightbigg 2 1 = π 6 (17 √ 17- 5 √ 5) 5. (a) Let E be the solid bounded by the paraboloids...
View Full Document

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern