Mixing
The volume of two vertically stacked, diagonally halved unit cubes
$begingroup$
How do I express the volume of the painted area using multiple integrals? The answer should be $1$, because the volume is $0.5 times 2 times 1 times 1=1$.
integration volume
edited Jan 31 at 7:40
Mohammad Zuhair Khan
1,6892625
asked Jan 31 at 7:33
KimKim
62
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add a comment |
$begingroup$
How do I express the volume of the painted area using multiple integrals? The answer should be $1$, because the volume is $0.5 times 2 times 1 times 1=1$.
integration volume
edited Jan 31 at 7:40
Mohammad Zuhair Khan
1,6892625
asked Jan 31 at 7:33
KimKim
62
$endgroup$
add a comment |
$begingroup$
How do I express the volume of the painted area using multiple integrals? The answer should be $1$, because the volume is $0.5 times 2 times 1 times 1=1$.
integration volume
edited Jan 31 at 7:40
Mohammad Zuhair Khan
1,6892625
asked Jan 31 at 7:33
KimKim
62
$endgroup$
How do I express the volume of the painted area using multiple integrals? The answer should be $1$, because the volume is $0.5 times 2 times 1 times 1=1$.
integration volume
integration volume
edited Jan 31 at 7:40
Mohammad Zuhair Khan
1,6892625
asked Jan 31 at 7:33
KimKim
62
edited Jan 31 at 7:40
Mohammad Zuhair Khan
1,6892625
asked Jan 31 at 7:33
KimKim
62
edited Jan 31 at 7:40
Mohammad Zuhair Khan
1,6892625
edited Jan 31 at 7:40
Mohammad Zuhair Khan
1,6892625
edited Jan 31 at 7:40
Mohammad Zuhair Khan
1,6892625
1,6892625
asked Jan 31 at 7:33
KimKim
62
asked Jan 31 at 7:33
KimKim
62
asked Jan 31 at 7:33
KimKim
62
62
add a comment |
add a comment |
2 Answers
2
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votes
$begingroup$
It is obvious that $0leq z leq 2$.
Let $f(t)$ be the area of the intersection of red area and $z=t$.
Then the volume is $int_0^2 f(t) dt$.
the intersection of red area and $z=t$ is expressed as follows:
$$1- frac{1}{2}t leq x leq 1, ~~ 0 leq y leq 1.$$
Thus $f(t)$ is
$$ int_0^1 int_{1- frac{1}{2}t}^1 dx dy.$$
Hence, the volume is
$$int_0^2 int_0^1 int_{1- frac{1}{2}t}^1 dx dy dt =
int_0^2 int_0^1 int_{1- frac{1}{2}z}^1 dx dy dz.
$$
answered Jan 31 at 8:24
Doyun NamDoyun Nam
66619
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add a comment |
$begingroup$
You can consider the cutting plane as graph of the function $f(x,y):=2x$ over the unit square $Q:=[0,1]times[0,1]$ in the $(x,y)$-plane. Your red set $B$ then has volume
$${rm vol}(B)=int_Q f(x,y)>{rm d}(x,y)=int_0^1int_0^1 2x >dy>dx=int_0^1 2x>dx=x^2biggr|_0^1=1 .$$
answered Jan 31 at 8:29
Christian BlatterChristian Blatter
176k8115328
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It is obvious that $0leq z leq 2$.
Let $f(t)$ be the area of the intersection of red area and $z=t$.
Then the volume is $int_0^2 f(t) dt$.
the intersection of red area and $z=t$ is expressed as follows:
$$1- frac{1}{2}t leq x leq 1, ~~ 0 leq y leq 1.$$
Thus $f(t)$ is
$$ int_0^1 int_{1- frac{1}{2}t}^1 dx dy.$$
Hence, the volume is
$$int_0^2 int_0^1 int_{1- frac{1}{2}t}^1 dx dy dt =
int_0^2 int_0^1 int_{1- frac{1}{2}z}^1 dx dy dz.
$$
answered Jan 31 at 8:24
Doyun NamDoyun Nam
66619
$endgroup$
add a comment |
$begingroup$
It is obvious that $0leq z leq 2$.
Let $f(t)$ be the area of the intersection of red area and $z=t$.
Then the volume is $int_0^2 f(t) dt$.
the intersection of red area and $z=t$ is expressed as follows:
$$1- frac{1}{2}t leq x leq 1, ~~ 0 leq y leq 1.$$
Thus $f(t)$ is
$$ int_0^1 int_{1- frac{1}{2}t}^1 dx dy.$$
Hence, the volume is
$$int_0^2 int_0^1 int_{1- frac{1}{2}t}^1 dx dy dt =
int_0^2 int_0^1 int_{1- frac{1}{2}z}^1 dx dy dz.
$$
answered Jan 31 at 8:24
Doyun NamDoyun Nam
66619
$endgroup$
add a comment |
$begingroup$
It is obvious that $0leq z leq 2$.
Let $f(t)$ be the area of the intersection of red area and $z=t$.
Then the volume is $int_0^2 f(t) dt$.
the intersection of red area and $z=t$ is expressed as follows:
$$1- frac{1}{2}t leq x leq 1, ~~ 0 leq y leq 1.$$
Thus $f(t)$ is
$$ int_0^1 int_{1- frac{1}{2}t}^1 dx dy.$$
Hence, the volume is
$$int_0^2 int_0^1 int_{1- frac{1}{2}t}^1 dx dy dt =
int_0^2 int_0^1 int_{1- frac{1}{2}z}^1 dx dy dz.
$$
answered Jan 31 at 8:24
Doyun NamDoyun Nam
66619
$endgroup$
It is obvious that $0leq z leq 2$.
Let $f(t)$ be the area of the intersection of red area and $z=t$.
Then the volume is $int_0^2 f(t) dt$.
the intersection of red area and $z=t$ is expressed as follows:
$$1- frac{1}{2}t leq x leq 1, ~~ 0 leq y leq 1.$$
Thus $f(t)$ is
$$ int_0^1 int_{1- frac{1}{2}t}^1 dx dy.$$
Hence, the volume is
$$int_0^2 int_0^1 int_{1- frac{1}{2}t}^1 dx dy dt =
int_0^2 int_0^1 int_{1- frac{1}{2}z}^1 dx dy dz.
$$
answered Jan 31 at 8:24
Doyun NamDoyun Nam
66619
answered Jan 31 at 8:24
Doyun NamDoyun Nam
66619
answered Jan 31 at 8:24
Doyun NamDoyun Nam
66619
answered Jan 31 at 8:24
Doyun NamDoyun Nam
66619
66619
add a comment |
add a comment |
$begingroup$
You can consider the cutting plane as graph of the function $f(x,y):=2x$ over the unit square $Q:=[0,1]times[0,1]$ in the $(x,y)$-plane. Your red set $B$ then has volume
$${rm vol}(B)=int_Q f(x,y)>{rm d}(x,y)=int_0^1int_0^1 2x >dy>dx=int_0^1 2x>dx=x^2biggr|_0^1=1 .$$
answered Jan 31 at 8:29
Christian BlatterChristian Blatter
176k8115328
$endgroup$
add a comment |
$begingroup$
You can consider the cutting plane as graph of the function $f(x,y):=2x$ over the unit square $Q:=[0,1]times[0,1]$ in the $(x,y)$-plane. Your red set $B$ then has volume
$${rm vol}(B)=int_Q f(x,y)>{rm d}(x,y)=int_0^1int_0^1 2x >dy>dx=int_0^1 2x>dx=x^2biggr|_0^1=1 .$$
answered Jan 31 at 8:29
Christian BlatterChristian Blatter
176k8115328
$endgroup$
add a comment |
$begingroup$
You can consider the cutting plane as graph of the function $f(x,y):=2x$ over the unit square $Q:=[0,1]times[0,1]$ in the $(x,y)$-plane. Your red set $B$ then has volume
$${rm vol}(B)=int_Q f(x,y)>{rm d}(x,y)=int_0^1int_0^1 2x >dy>dx=int_0^1 2x>dx=x^2biggr|_0^1=1 .$$
answered Jan 31 at 8:29
Christian BlatterChristian Blatter
176k8115328
$endgroup$
You can consider the cutting plane as graph of the function $f(x,y):=2x$ over the unit square $Q:=[0,1]times[0,1]$ in the $(x,y)$-plane. Your red set $B$ then has volume
$${rm vol}(B)=int_Q f(x,y)>{rm d}(x,y)=int_0^1int_0^1 2x >dy>dx=int_0^1 2x>dx=x^2biggr|_0^1=1 .$$
answered Jan 31 at 8:29
Christian BlatterChristian Blatter
176k8115328
answered Jan 31 at 8:29
Christian BlatterChristian Blatter
176k8115328
answered Jan 31 at 8:29
Christian BlatterChristian Blatter
176k8115328
answered Jan 31 at 8:29
Christian BlatterChristian Blatter
176k8115328
176k8115328
add a comment |
add a comment |
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