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
Unformatted text preview: the surface z = xy that lies within the cylinder x 2 + y 2 = 1 . Solution. With f ( x, y ) = xy, the surface area formula has f x = y, f y = x, so p y 2 + x 2 + 1 and Area(S) = ZZ D p y 2 + x 2 + 1 dA, where D is the region inside the circle x 2 + y 2 = 1 . Switching to polar coords, we get the iterated integral Z 2 Z 1 p r 2 + 1 r drd. Problem 15.7.3. Evaluate the iterated integral Z 1 Z z Z x + z 6 xz dydxdz. Solution. The rst inside integral has the anti-derivative Z 6 xz dy = 6 xyz, so the denite integral is Z x + z 6 xz dy = [6 xyz ] y = x + z y =0 = 6 x ( x + z ) . From this point, the problem is a double iterated integral, with value =1. 3 Finally, we covered problems 15.7.9 (used for #35) then 15.8.10 and 15.8.18 all of which are past the syllabus for Exam 2, but will be needed in Ch. 16....
View Full Document
- Spring '06