Calculating volume of a set $K={(x,y,z)^T in mathbb{R}^3:x^2+y^2+z^2 le 4, x^2+y^2 le 2x}$
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I want to to calculcate the volume, that is $lambda_3$, of $K={(x,y,z)^T in mathbb{R}^3:x^2+y^2+z^2 le 4, x^2+y^2 le 2x}$.
I guess I need to evaluate a triple integral for this but I don't know how to this task.
calculus real-analysis integration analysis lebesgue-integral
asked Nov 17 at 16:25
conrad
657
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up vote
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down vote
favorite
I want to to calculcate the volume, that is $lambda_3$, of $K={(x,y,z)^T in mathbb{R}^3:x^2+y^2+z^2 le 4, x^2+y^2 le 2x}$.
I guess I need to evaluate a triple integral for this but I don't know how to this task.
calculus real-analysis integration analysis lebesgue-integral
asked Nov 17 at 16:25
conrad
657
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I want to to calculcate the volume, that is $lambda_3$, of $K={(x,y,z)^T in mathbb{R}^3:x^2+y^2+z^2 le 4, x^2+y^2 le 2x}$.
I guess I need to evaluate a triple integral for this but I don't know how to this task.
calculus real-analysis integration analysis lebesgue-integral
asked Nov 17 at 16:25
conrad
657
I want to to calculcate the volume, that is $lambda_3$, of $K={(x,y,z)^T in mathbb{R}^3:x^2+y^2+z^2 le 4, x^2+y^2 le 2x}$.
I guess I need to evaluate a triple integral for this but I don't know how to this task.
calculus real-analysis integration analysis lebesgue-integral
calculus real-analysis integration analysis lebesgue-integral
asked Nov 17 at 16:25
conrad
657
asked Nov 17 at 16:25
conrad
657
asked Nov 17 at 16:25
conrad
657
asked Nov 17 at 16:25
conrad
657
asked Nov 17 at 16:25
conrad
657
657
add a comment |
add a comment |
1 Answer
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In this type of problems, try to make a sketch of the volume. The first inequality is the interior of the sphere with radius $2$, centered at the origin.
If you move $2x$ to the other side in the second inequality, then complete the square, you get that this is the interior of the cylinder of radius $1$, with axis parallel to $z$, and centered on $x=1$.
So one way to set up the calculation is to look at the $z=0$ plane. The integration range in this plane is the circle with center at $x=1$, $y=0$ and radius $1$. So the limits for $x$ are $0$ to $2$, and the limits for $y$ are from $-sqrt{2x-x^2}$ to $+sqrt{2x-x^2}$. The limits for $z$ you get from the first inequality, $z$ varies between $pmsqrt{4-y^2-x^2}$. Therefore
$$V=int_0^2dxint_{-sqrt{2x-x^2}}^{sqrt{2x-x^2}}dyint_{-sqrt{4-y^2-x^2}}^sqrt{4-y^2-x^2}dz$$
answered Nov 17 at 16:46
Andrei
10.2k21025
How did you get the limits for $y$? Did you forget the $sqrt{}$?
– conrad
Nov 17 at 16:58
Yes I did. I will fix it
– Andrei
Nov 17 at 16:59
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
In this type of problems, try to make a sketch of the volume. The first inequality is the interior of the sphere with radius $2$, centered at the origin.
If you move $2x$ to the other side in the second inequality, then complete the square, you get that this is the interior of the cylinder of radius $1$, with axis parallel to $z$, and centered on $x=1$.
So one way to set up the calculation is to look at the $z=0$ plane. The integration range in this plane is the circle with center at $x=1$, $y=0$ and radius $1$. So the limits for $x$ are $0$ to $2$, and the limits for $y$ are from $-sqrt{2x-x^2}$ to $+sqrt{2x-x^2}$. The limits for $z$ you get from the first inequality, $z$ varies between $pmsqrt{4-y^2-x^2}$. Therefore
$$V=int_0^2dxint_{-sqrt{2x-x^2}}^{sqrt{2x-x^2}}dyint_{-sqrt{4-y^2-x^2}}^sqrt{4-y^2-x^2}dz$$
answered Nov 17 at 16:46
Andrei
10.2k21025
How did you get the limits for $y$? Did you forget the $sqrt{}$?
– conrad
Nov 17 at 16:58
Yes I did. I will fix it
– Andrei
Nov 17 at 16:59
add a comment |
up vote
0
down vote
In this type of problems, try to make a sketch of the volume. The first inequality is the interior of the sphere with radius $2$, centered at the origin.
If you move $2x$ to the other side in the second inequality, then complete the square, you get that this is the interior of the cylinder of radius $1$, with axis parallel to $z$, and centered on $x=1$.
So one way to set up the calculation is to look at the $z=0$ plane. The integration range in this plane is the circle with center at $x=1$, $y=0$ and radius $1$. So the limits for $x$ are $0$ to $2$, and the limits for $y$ are from $-sqrt{2x-x^2}$ to $+sqrt{2x-x^2}$. The limits for $z$ you get from the first inequality, $z$ varies between $pmsqrt{4-y^2-x^2}$. Therefore
$$V=int_0^2dxint_{-sqrt{2x-x^2}}^{sqrt{2x-x^2}}dyint_{-sqrt{4-y^2-x^2}}^sqrt{4-y^2-x^2}dz$$
answered Nov 17 at 16:46
Andrei
10.2k21025
How did you get the limits for $y$? Did you forget the $sqrt{}$?
– conrad
Nov 17 at 16:58
Yes I did. I will fix it
– Andrei
Nov 17 at 16:59
add a comment |
up vote
0
down vote
up vote
0
down vote
In this type of problems, try to make a sketch of the volume. The first inequality is the interior of the sphere with radius $2$, centered at the origin.
If you move $2x$ to the other side in the second inequality, then complete the square, you get that this is the interior of the cylinder of radius $1$, with axis parallel to $z$, and centered on $x=1$.
So one way to set up the calculation is to look at the $z=0$ plane. The integration range in this plane is the circle with center at $x=1$, $y=0$ and radius $1$. So the limits for $x$ are $0$ to $2$, and the limits for $y$ are from $-sqrt{2x-x^2}$ to $+sqrt{2x-x^2}$. The limits for $z$ you get from the first inequality, $z$ varies between $pmsqrt{4-y^2-x^2}$. Therefore
$$V=int_0^2dxint_{-sqrt{2x-x^2}}^{sqrt{2x-x^2}}dyint_{-sqrt{4-y^2-x^2}}^sqrt{4-y^2-x^2}dz$$
answered Nov 17 at 16:46
Andrei
10.2k21025
In this type of problems, try to make a sketch of the volume. The first inequality is the interior of the sphere with radius $2$, centered at the origin.
If you move $2x$ to the other side in the second inequality, then complete the square, you get that this is the interior of the cylinder of radius $1$, with axis parallel to $z$, and centered on $x=1$.
So one way to set up the calculation is to look at the $z=0$ plane. The integration range in this plane is the circle with center at $x=1$, $y=0$ and radius $1$. So the limits for $x$ are $0$ to $2$, and the limits for $y$ are from $-sqrt{2x-x^2}$ to $+sqrt{2x-x^2}$. The limits for $z$ you get from the first inequality, $z$ varies between $pmsqrt{4-y^2-x^2}$. Therefore
$$V=int_0^2dxint_{-sqrt{2x-x^2}}^{sqrt{2x-x^2}}dyint_{-sqrt{4-y^2-x^2}}^sqrt{4-y^2-x^2}dz$$
answered Nov 17 at 16:46
Andrei
10.2k21025
edited Nov 17 at 17:00
answered Nov 17 at 16:46
Andrei
10.2k21025
answered Nov 17 at 16:46
Andrei
10.2k21025
answered Nov 17 at 16:46
Andrei
10.2k21025
10.2k21025
How did you get the limits for $y$? Did you forget the $sqrt{}$?
– conrad
Nov 17 at 16:58
Yes I did. I will fix it
– Andrei
Nov 17 at 16:59
add a comment |
How did you get the limits for $y$? Did you forget the $sqrt{}$?
– conrad
Nov 17 at 16:58
Yes I did. I will fix it
– Andrei
Nov 17 at 16:59
How did you get the limits for $y$? Did you forget the $sqrt{}$?
– conrad
Nov 17 at 16:58
How did you get the limits for $y$? Did you forget the $sqrt{}$?
– conrad
Nov 17 at 16:58
Yes I did. I will fix it
– Andrei
Nov 17 at 16:59
Yes I did. I will fix it
– Andrei
Nov 17 at 16:59
add a comment |
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