{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Math252Sols4F07

# Math252Sols4F07 - QUIZ 4(CHAPTER 17 SOLUTIONS MATH 252 FALL...

This preview shows pages 1–4. Sign up to view the full content.

QUIZ 4 (CHAPTER 17) SOLUTIONS MATH 252 – FALL 2007 – KUNIYUKI SCORED OUT OF 125 POINTS MULTIPLIED BY 0.84 105% POSSIBLE 1) Reverse the order of integration, and evaluate the resulting double integral: cos 1 + y 4 ( ) dy dx x 3 3 0 27 . Give a simplified exact answer; do not approximate. Sketch the region of integration. (20 points) Sketch the region of integration, R . It may help to rewrite the double integral with labels: cos 1 + y 4 ( ) dy dx y = x 3 y = 3 x = 0 x = 27 We aim / fix x , shoot y , and slide x . Reverse the order of integration. Solve y = x 3 for x : x = y 3

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

View Full Document
We aim y , shoot x , and slide y . cos 1 + y 4 ( ) dx dy x = 0 x = y 3 y = 0 y = 3 = x cos 1 + y 4 ( ) x = 0 x = y 3 dy y = 0 y = 3 = y 3 cos 1 + y 4 ( ) 0 ( ) dy 0 3 = y 3 cos 1 + y 4 ( ) dy 0 3 We now perform a standard u -substitution. Let u = 1 + y 4 du = 4 y 3 dy y 3 dy = 1 4 du You can also “Compensate”: y 3 cos 1 + y 4 ( ) dy 0 3 = 1 4 4 y 3 cos 1 + y 4 ( ) dy 0 3 Change the limits of integration: y = 0 u = 1 + 0 ( ) 4 u = 1 y = 3 u = 1 + 3 ( ) 4 u = 82 = cos u 1 4 du u = 1 u = 82 = 1 4 sin u 1 82 = 1 4 sin82 sin1 ( ) Note: This is about 0.13206. ( )
2) Let R be the region in the xy -plane that is bounded by the rectangle with vertices 1, 3 ( ) , 7, 3 ( ) , 7, 5 ( ) , and 1, 5 ( ) . Set up a double integral for the surface area of the portion of the graph of 9 x 2 + 4 y 2 + z 2 = 1000 z > 0 ( ) that lies over R . Make sure your double integral is as detailed as possible; do not leave in generic notation like R or f . Also, do not leave dA in your final answer; break it down into d variable ( ) d variable ( ) . Do not evaluate. (14 points) Observe that the graph of 9 x 2 + 4 y 2 + z 2 = 1000 is an ellipsoid. With the restriction z > 0 ( ) , we only consider [part of] the upper half. Solve 9 x 2 + 4 y 2 + z 2 = 1000 for z : 9 x 2 + 4 y 2 + z 2 = 1000 z > 0 ( ) z 2 = 1000 9 x 2 4 y 2 z > 0 ( ) z = 1000 9 x 2 4 y 2 Let f x , y ( ) = 1000 9 x 2 4 y 2 . Its domain contains R . S = 1 + f x x , y ( ) [ ] 2 + f y x , y ( ) [ ] 2 dA R ∫∫ f x , y ( ) = 1000 9 x 2 4 y 2 ( ) 1 2 f x x , y ( ) = 1 2 1000 9 x 2 4 y 2 ( ) 1 2 18 x ( ) = 9 x 1000 9 x 2 4 y 2 f y x , y ( ) = 1 2 1000 9 x 2 4 y 2 ( ) 1 2 8 y ( ) = 4 y 1000 9 x 2 4 y 2 S = 1 + 9 x 1000 9 x 2 4 y 2 2 + 4 y 1000 9 x 2 4 y 2 2 dy dx 3 5 1 7 or S = 1 + 81 x 2 1000 9 x 2 4 y 2 + 16 y 2 1000 9 x 2 4 y 2 dy dx 3 5 1 7 or S = 1 + 81 x 2 + 16 y 2 1000 9 x 2 4 y 2 dy dx 3 5

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.

{[ snackBarMessage ]}