# ReviewHW13-solutions - huynh(lth436 ReviewHW13...

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huynh (lth436) – ReviewHW13 – gilbert – (56690) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 0.0 points Evaluate the integral I = integraldisplay S 2 xyz dS when S is the part of the plane 2 z = 2 x + 2 y that lies above the rectangle [0 , 1] × [0 , 1] in the xy -plane. 1. I = 1 correct 2. I = 2 3. I = 7 4 4. I = 3 2 5. I = 5 4 Explanation: The surface area element dS for the graph of any function z = f ( x, y ) is given by dS = ( f 2 x + f 2 y + 1) 1 / 2 dxdy . But when z = f ( x, y ) = 1 x 2 + y , we see that f 2 x + f 2 y + 1 = 1 4 + 1 + 1 = 9 4 . On the other hand, on S , 2 xyz = 2 xyf ( x, y ) = 2 xy x 2 y + 2 xy 2 . Thus, as a repeated integral, I = 3 2 integraldisplay 1 0 parenleftBig integraldisplay 1 0 (2 xy x 2 y + 2 xy 2 ) dy parenrightBig dx . Now integraldisplay 1 0 (2 xy x 2 y + 2 xy 2 ) dy = bracketleftBig xy 2 1 2 x 2 y 2 + 2 3 xy 3 bracketrightBig 1 0 , and so I = 3 2 integraldisplay 1 0 parenleftBig x 1 2 x 2 + 2 3 x parenrightBig dx = 3 2 bracketleftBig 1 2 x 2 1 6 x 3 + 1 3 x 2 bracketrightBig 1 0 . Consequently, I = 3 2 parenleftBig 1 2 1 6 + 1 3 parenrightBig = 1 . 002 0.0 points Evaluate the integral I = integraldisplay S yz dS when S is the part of the plane x + 2 y + 2 z = 0 shown in
huynh (lth436) – ReviewHW13 – gilbert – (56690) 2 enclosed by the cylinder x 2 + y 2 = 4 . 1. I = 3 π 2. I = 3 π 3. I = 6 π 4. I = 0 5. I = 6 π correct Explanation: The surface area element dS for the graph of any function z = f ( x, y ) is given by dS = ( f 2 x + f 2 y + 1) 1 / 2 dxdy . But when z = f ( x, y ) = x 2 y , we see that f 2 x + f 2 y + 1 = 1 4 + 1 + 1 = 9 4 . On the other hand, on S , yz = yf ( x, y ) = parenleftBig xy 2 + y 2 parenrightBig . Thus I = 3 2 integraldisplay integraldisplay D parenleftBig xy 2 + y 2 parenrightBig dxdy , where D = { ( x, y ) : x 2 + y 2 4 } . Because of the rotational symmetry, we’ll use polar coordinates to evaluate I . For then I can be written as 3 2 integraldisplay 2 0 integraldisplay 2 π 0 r 2 parenleftBig 1 2 cos θ sin θ + sin 2 θ parenrightBig r dθdr. Now 1 2 cos θ sin θ + sin 2 θ = 1 4 sin 2 θ + 1 2 (1 cos 2 θ ) , and so integraldisplay 2 π 0 parenleftBig 1 2 cos θ sin θ + sin 2 θ parenrightBig = 1 4 integraldisplay 2 π 0 (sin 2 θ + 2 2 cos 2 θ ) = π . Thus I = 3 2 π integraldisplay 2 0 r 3 dr = 3 2 . 16 4 = 6 π . 003 0.0 points Evaluate the integral I = 1 4 integraldisplay S dS when S is the surface given parametrically by Φ ( u, v ) = (2 uv, u + v, u v ) for u 2 + v 2 4. 1. I = 4 π 2. I = 10 3 π 3. I = 13 3 π correct 4. I = 14 3 π 5. I = 11 3 π Explanation: When S is parametrized by Φ ( u, v ) = (2 uv, u + v, u v )
huynh (lth436) – ReviewHW13 – gilbert – (56690) 3 for u 2 + v 2 4, then I = 1 4 integraldisplay integraldisplay D bardbl T u × T v bardbl dudv , where D = { ( u, v ) : u 2 + v 2 4 } . Now T u = Φ ∂u = (2 v, 1 , 1) , while T v = Φ ∂v = (2 u, 1 , 1) . In this case, T u × T v = vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle i j k 2 v 1 1 2 u 1 1 vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle = 2 i + 2( u + v ) j + 2( v u ) k . Thus bardbl T u × T v bardbl = 2(1 + ( u + v ) 2 + ( v u ) 2 ) 1 / 2 = 2(1 + 2( u 2 + v 2 )) 1 / 2 . So, finally, we arrive at I = 1 2 integraldisplay integraldisplay D (1 + 2( u 2 + v 2 )) 1 / 2 dudv . Because of the rotational symmetry, we’ll use polar coordinates with u = r cos θ , v = r sin θ , to evaluate I