{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

tutorial7_sol

tutorial7_sol - dx = 1 6 sin x 2 ± ± ± 3 = sin 9 6 3 The...

This preview shows page 1. Sign up to view the full content.

MATH1211/10-11(2)/Tu7sol/TNK Department of Mathematics, The University of Hong Kong MATH1211 Multivariable Calculus (2010-11 2nd semester) Solution to Tutorial 7 Note : Some solutions given below are outline only. You may need to give more details in your solution. 1. The two regions of the two integrals are shown as in the ﬁgure at right. By changing the order of integration, we combine the two regions together and rewrite the sum of integrals as one single integral and evaluate: Z 1 0 Z 2 - y y sin xdxdy = Z 1 0 - cos x ± ± ± x =2 - y x = y dy = Z 1 0 ( cos y - cos(2 - y ) ) dy = ( sin y + sin(2 - y ) ) ± ± ± 1 0 = 2 sin 1 - sin 2 . 2. Note that cos( x 2 ) does not have an anti-derivative that can be written in terms of elementary functions. Instead we reverse the order of integration (see the region of integration in the ﬁgure at right) to obtain: Z 1 0 Z 3 3 y cos( x 2 ) dxdy = Z 3 0 Z 1 3 x 0 cos( x 2 ) dy dx = Z 3 0 1 3 x cos( x 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ) dx = 1 6 sin( x 2 ) ± ± ± 3 = sin 9 6 . 3. The region of integration is shown in ﬁgure at right. Z 1 Z 2 Z x 2 f ( x,y,z ) dz dxdy = Z 2 Z 1 Z x 2 f ( x,y,z ) dz dy dx = Z 2 Z x 2 Z 1 f ( x,y,z ) dy dz dx = Z 4 Z 2 √ z Z 1 f ( x,y,z ) dy dxdz = Z 1 Z 4 Z 2 √ z f ( x,y,z ) dxdz dy = Z 4 Z 1 Z 2 √ z f ( x,y,z ) dxdy dz. 4. The answer is True. Reason: the region D of integration, which is an elliptical disc centred at (2 , 0) with major axis on the x-axis, is symmetric about the line y = 0. On the other hand, y 3 is an odd function because (-y ) 3 =-y 3 . Hence RR D y 3 dxdy = 0, which means that ZZ D ( y 3 + 1) dxdy = ZZ D y 3 dxdy + ZZ D 1 dxdy = ZZ D 1 dxdy = area of D. 1...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online