Question

Express the integral ∭

E
​
f(x,y,z)dV as an iterated integral in six different ways, where E is the solid bounded by z=0,x=0,z=y−5x and y=15. 1. ∫
a
b
​
∫
g
1
​
(x)
g
2
​
(x)
​
∫
h
1
​
(x,y)
h
2
​
(x,y)
​
f(x,y,z)dzdydx a=−
b=
− g
1
​
(x)=
g
2
​

​
(x)= h
1
​
(x,y)=−h
2
​
(x,y)= 2. ∫
a
b
​
∫
g
1
​
(y)
g
2
​
(y)
​
∫
h
1
​
(x,y)
h
2
​
(x,y)
​
f(x,y,z)dzdxdy a=b= g
1
​
(y)=−g
2
​
(y)= h
1
​
(x,y)=h
2
​
(x,y)= 3. ∫
a
b
​
∫
g
1
​
(z)
g
2
​
(z)
​
∫
h
1
​
(y,z)
h
2
​
(y,z)
​
f(x,y,z)dxdydz a=b= g
1
​
(z)=
g
2
​
(z)
​
= h
1
​
(y,z)=h
2
​
(y,z)= 4. ∫
a
b
​
∫
g
1
​
(y)
g
2
​
(y)
​
∫
h
1
​
(y,z)
h
2
​
(y,z)
​
f(x,y,z)dxdzdy a=b= g
1
​
(y)=−g
2
​
(y)= h
1
​
(y,z)=h
2
​
(y,z)= 5. ∫
a
b
​
∫
g
1
​
(x)
g
2
​
(x)
​
∫
h
1
​
(x,z)
h
2
​
(x,z)
​
f(x,y,z)dydzdx a=b= g
1
​
(x)=−
g
2
​
(x)
​
= h
1
​
(x,z)=h
2
​
(x,z)= 6. ∫
a
b
​
∫
g
1
​
(z)
g
2
​
(z)
​
∫
h
1
​
(x,z)
h
2
​
(x,z)
​
f(x,y,z)dydxdz a=b= g
1
​
(z)=−
g
2
​
(z)
​
= h
1
​
(x,z)=h
2
​
(x,z)=

Answer 1

Explanation:

We are given the solid E bounded by z=0, x=0, z=y−5x and y=15.

To express the integral ∭ E f(x,y,z)dV as an iterated integral in six different ways, we can use the bounds of integration for each variable x, y, and z.

Method 1: Integrating with respect to z first

The bounds of z are from 0 to y-5x. The bounds of y are from x+5x to 15. The bounds of x are from 0 to 3.

∫0^3 ∫x+5x^15 ∫0^(y-5x) f(x,y,z) dz dy dx

Method 2: Integrating with respect to z last

The bounds of z are from 0 to y-5x. The bounds of y are from x+5x to 15. The bounds of x are from 0 to 3.

∫x=0^3 ∫y=x+5x^15 ∫z=0^(y-5x) f(x,y,z) dz dy dx

Method 3: Integrating with respect to y first

The bounds of y are from 5x to 15. The bounds of x are from 0 to (y/5).

∫0^3 ∫5x^15 ∫0^(y-5x) f(x,y,z) dz dy dx

Method 4: Integrating with respect to y last

The bounds of y are from 5x to 15. The bounds of x are from 0 to (y/5).

∫5x^15 ∫0^(y/5) ∫z=0^(y-5x) f(x,y,z) dz dx dy

Method 5: Integrating with respect to x first

The bounds of x are from 0 to (y/5). The bounds of y are from 5x to 15. The bounds of z are from 0 to y-5x.

∫0^3 ∫5x^15 ∫0^(y-5x) f(x,y,z) dz dy dx

Method 6: Integrating with respect to x last

The bounds of x are from 0 to (y/5). The bounds of y are from 5x to 15. The bounds of z are from 0 to y-5