= 4 and the plane y3. A region in the first octant is bounded by the cylinder x² + z²up the integrals for its volume in the order (a) dx dy dz, (b) dz dx dy, and (c) dy dz dx.=3. Set

∫∫∫dxdydz=∫02∫03∫04-z2dxdydz=∫02∫034-z2dydz=∫02y034-z2dz=3∫024-z2dz=32z4-z2+4sin-1z202…

Answered: 3. A region in the first octant is…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Related questions

Question

= 4 and the plane y
3. A region in the first octant is bounded by the cylinder x² + z²
up the integrals for its volume in the order (a) dx dy dz, (b) dz dx dy, and (c) dy dz dx.
=
3. Set

Expert Solution

Step 1: Finding the volume

$\int_{}^{} \int_{}^{} \int_{}^{} d x d y d z = \int_{0}^{2} \int_{0}^{3} \int_{0}^{\sqrt{4 - z^{2}}} d x d y d z = \int_{0}^{2} \int_{0}^{3} \sqrt{4 - z^{2}} d y d z = \int_{0}^{2} {[y]}_{0}^{3} \sqrt{4 - z^{2}} d z = 3 \int_{0}^{2} \sqrt{4 - z^{2}} d z = \frac{3}{2} {[z \sqrt{4 - z^{2}} + 4 \sin^{- 1} (\frac{z}{2})]}_{0}^{2} [∵ \int_{}^{} \sqrt{a^{2} - x^{2}} d x = \frac{x \sqrt{a^{2} - x^{2}}}{2} + \frac{a^{2}}{2} \sin^{- 1} (\frac{x}{a})] = \frac{3}{2} \times 4 \sin^{- 1} (1) = 6 \times \frac{π}{2} = 3 π$