Step by step, checkpoint 5.12 ## Example 5.17### Finding the Volume of a TetrahedronFind the volume of the solid bounded by the planes $ x = 0 $, $ y = 0 $, $ z = 0 $, and $ 2x + 3y + z = 6 $.[Show Solution]---## Checkpoint 5.12Find the volume of the solid bounded above by $ f(x, y) = 10 - 2x + y $ over the region enclosed by the curves $ y = 0 $ and $ y = e^x $, where $ x $ is in the interval $[0, 1]$.---Finding the area of a rectangular region is easy, but finding the area of a nonrectangular region is not so easy. As we have seen, we can use double integrals to find a rectangular area. As a matter of fact, this comes in very handy for finding the area of a general nonrectangular region, as stated in the next definition.

We are given:f(x, y) = 10 - 2x + y, y = 0, y = ex0≤x≤1We can evaluate this integral as I = ∫01∫0exf(x,y) dydx = ∫01∫0ex10 - 2x + y dydx = ∫01[(10 - 2x) y + y22]0ex dx = ∫01[(10 - 2x) ex + (ex)22]-0 dx = ∫01[(10 - 2x) ex + e2x2]dx = ∫0110ex - 2xex + e2x2 dxOn solving further, we get, =10[ex]01-2[xex-ex]01 +12[e2x2]01 =10[e-1] - 2[(e-e)-(0-1)] + 12[e22-12]= 10e - 10 -2 +e24-14=e24 +10e-494 cubic units This is the required volume of the solid bounded by the given region.

Answered: Find the volume of the solid bounded…

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Find the volume of the solid bounded above by f (x, y) = 10-2x + y over the region enclosed by the curves y = 0 and y = e, where x is in the interval [0, 1]. %3D %3D %3D

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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Question

Step by step, checkpoint 5.12

## Example 5.17

### Finding the Volume of a Tetrahedron

Find the volume of the solid bounded by the planes $ x = 0 $, $ y = 0 $, $ z = 0 $, and $ 2x + 3y + z = 6 $.

[Show Solution]

---

## Checkpoint 5.12

Find the volume of the solid bounded above by $ f(x, y) = 10 - 2x + y $ over the region enclosed by the curves $ y = 0 $ and $ y = e^x $, where $ x $ is in the interval $[0, 1]$.

---

Finding the area of a rectangular region is easy, but finding the area of a nonrectangular region is not so easy. As we have seen, we can use double integrals to find a rectangular area. As a matter of fact, this comes in very handy for finding the area of a general nonrectangular region, as stated in the next definition.

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