If you could please help me solve and show your work! I would be so grateful! Thank you so very much! The surface of an island is defined by the following function over the region on which the function is nonnegative. Find the volume V of the island shown to the right.z=25-V+yz= 25 - Vx +y?Set up the double integral, in polar coordinates, that is used to find the volume.(Type exact answers.)

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Description: If you could please help me solve and show your work! I would be so grateful! Thank you so very much! The surface of an island is defined by the following function over the region on which the function is nonnegative. Find the volume V of the island

Answered: please help me solve and show your work

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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If you could please help me solve and show your work! I would be so grateful! Thank you so very much!

The surface of an island is defined by the following function over the region on which the function is nonnegative. Find the volume V of the island shown to the right.
z=25-V+y
z= 25 - Vx +y?
Set up the double integral, in polar coordinates, that is used to find the volume.
(Type exact answers.)

Expert Solution

Step 1

Given

$Z = 25 - \sqrt{x^{2} + y^{2}}$

Let W be the solid region bounded by the surfaces $Z = 25 - \sqrt{x^{2} + y^{2}}$ and the domain $D = {(x, y) : x^{2} + y^{2} \leq 25^{2}}$

Then $W = \{(x, y, z) : 0 < z < 25 - \sqrt{x^{2} + y^{2}}, - \sqrt{625 + x^{2}} < y < \sqrt{625 + x^{2}}, - 25 < x < 25\}$

and

$D = \{(x, y) : - \sqrt{625 + x^{2}} < y < \sqrt{625 + x^{2}}, - 25 < x < 25\}$

Step 2

The volume of the solid region W is given by the triple integral

$V = \int_{- 25}^{25} \int_{- \sqrt{625 - x^{2}}}^{\sqrt{625 - x^{2}}} \int_{0}^{25 - \sqrt{x^{2} + y^{2}}} d z d y d x$ , where $f (x, y, z) = 25 - \sqrt{x^{2} + y^{2}}$

To convert this volume integral into polar coordinates Put $x = r \cos θ, y = r \sin θ, t h e n r = \sqrt{x^{2} + y^{2}}$

since the base of the cone is circular therefore $0 \leq θ < 2 π and 0 < r < 25$

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