s07wk09 - Math 23 B Dodson Week 9 Homework 15.2 iterated...

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

View Full Document Right Arrow Icon
Math 23 B. Dodson Week 9 Homework: 15.2 iterated integrals [re-print of week 8 example] 15.3 general regions 15.4 polar coords Week 10 Homework: 15.5 Applications of Double Integrals [Omit #23] . [End of Exam 2 syllabus, Thurs March 29] 15.6 Surface Area [Omit #12; End of Hw 10, due Fri. March 30] 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,
Image of page 1

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

View Full Document Right 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 that approximate the region over R and under the graph of 1 + 4 xy. In particular, our calculation shows that this region has volume 10 (cubic units).
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern