Math  /  Calculus

Question7. Let E={(x,y,z)x+y+z1,x0,y0,z0}E=\{(x, y, z) \mid x+y+z \leq 1, x \geq 0, y \geq 0, z \geq 0\}. Show that EezdV=e52\iiint_{E} e^{z} d V=e-\frac{5}{2}

Studdy Solution

STEP 1

What is this asking? We need to calculate the triple integral of eze^z over a tetrahedral region defined by some inequalities, and show it equals e52e - \frac{5}{2}. Watch out! Setting up the integration limits correctly is crucial here!
We need to make sure the limits reflect the tetrahedron's shape.

STEP 2

1. Define the integration limits
2. Perform the triple integration

STEP 3

Alright, let's **define our region** EE.
We have x+y+z1x + y + z \leq 1 with x0x \geq 0, y0y \geq 0, and z0z \geq 0.
This describes a tetrahedron in the first octant!

STEP 4

Let's **set up the integral** in the order dzdydxdz\,dy\,dx.
Since x+y+z1x + y + z \leq 1, we know zz goes from 00 to 1xy1 - x - y.

STEP 5

Now for the yy limits.
When z=0z = 0, we have x+y1x + y \leq 1, so yy goes from 00 to 1x1 - x.

STEP 6

Finally, the xx limits.
Since x0x \geq 0 and when yy and zz are both zero, we have x1x \leq 1, so xx goes from 00 to 11.

STEP 7

Let's **write out our triple integral** with the limits we just found: EezdV=0101x01xyezdzdydx \iiint_E e^z \,dV = \int_0^1 \int_0^{1-x} \int_0^{1-x-y} e^z \,dz\,dy\,dx

STEP 8

**Innermost integral (with respect to** zz): 01xyezdz=[ez]01xy=e1xye0=e1xy1 \int_0^{1-x-y} e^z \,dz = \left[ e^z \right]_0^{1-x-y} = e^{1-x-y} - e^0 = e^{1-x-y} - 1

STEP 9

**Middle integral (with respect to** yy): \begin{align*} \int_0^{1-x} (e^{1-x-y} - 1) \,dy &= \left[ -e^{1-x-y} - y \right]_0^{1-x} \\ &= (-e^0 - (1-x)) - (-e^{1-x} - 0) \\ &= -1 - 1 + x + e^{1-x} = x + e^{1-x} - 2\end{align*}

STEP 10

**Outermost integral (with respect to** xx): \begin{align*} \int_0^1 (x + e^{1-x} - 2) \,dx &= \left[ \frac{x^2}{2} - e^{1-x} - 2x \right]_0^1 \\ &= \left( \frac{1}{2} - e^0 - 2 \right) - \left( 0 - e^1 - 0 \right) \\ &= \frac{1}{2} - 1 - 2 + e = e - \frac{5}{2}\end{align*}

STEP 11

We've shown that the triple integral evaluates to e52e - \frac{5}{2}, as required!

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