Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 15: Multiple Integrals - Practice Exercises - Page 933: 13



Work Step by Step

We calculate the integral as follows: Area; $$A=\int_{-2}^{0} \int_{ 2x+4}^{4-x^2} dy dx \\=\int_{-2}^{0} (-x^2 -2x) dx \\=- \int_{-2}^{0} x^2 \ dx - 2 \int_{-2}^{0} x \ dx \\=-\dfrac{1}{3} \times [ x^3]_{-2}^0 - [ x^2]_{-2}^0 \\=-[ 0-\dfrac{(-2)^3}{3}] - [0- (-2)^2] \\=\dfrac{4}{3}$$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.