s10wk08 - Math 23 B. Dodson Week 8 Homework: 15.2 iterated...

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

View Full Document Right Arrow Icon
Math 23 B. Dodson Week 8 Homework: 15.2 iterated integrals [re-print of week 7 example] 15.3 general regions 15.4 polar coords Week 9 Homework: 15.5 Applications of Double Integrals 15.6 Triple integrals 15.7 Cylindrical Coordinates; 15.8 Spherical Coordinates [End of Exam 2 syllabus, Thurs April 8] Problem 15.2.3: Evaluate the iterated integral Z 3 1 Z 1 0 (1 + 4 xy ) dxdy. Solution: We start with the inside integral: Z 1 0 (1 + 4 xy ) dx and find an anti-derivative (under ∂x ) for 1 + 4 xy. Holding y constant, we find ∂x ( x + 2 x 2 y ) = 1 + 4 xy, so Z 1 0 (1 + 4 xy ) dx = £ x + 2 x 2 y / 1 0 = (1 + 2 y ) - (0 + 0) = 1 + 2 y. We replace the inside integral with this value,
Background image of page 1

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

View Full DocumentRight Arrow Icon
2 giving the outside integral: Z 3 1 (1 + 2 y ) dy = £ y + y 2 / 3 1 = (3 + 9) - (1 + 1) = 10 . We recall that this calculates the value of the double integral ZZ R (1 + 4 xy ) dA, over the rectangle R = [0 , 1] × [1 , 3] , defined as the limit of approximating sums that can be regarded as sums of volumes of rectangular solids
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 3

s10wk08 - Math 23 B. Dodson Week 8 Homework: 15.2 iterated...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online