Looking for where I went wrong: Finding the volume of a solid that lies within both a cylinder and sphere
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I'm currently working on this question:
Find the volume of the solid that lies within both the cylinder $x^2+y^2=1$ and the sphere $x^2+y^2+z^2=4$.
I decided to use polar coordinates so that the cylinder equation becomes $r^2=1$ and the sphere becomes $r^2+z^2=4$.
Solving for $z$, I get the inequality $-sqrt{4-r^2}leq zleq sqrt{4-r^2}$. Since I know what $r^2$ is, I plug that in to get the inequality where $z$ is between $-sqrt{3}$ and $sqrt{3}$. Combining that to make a triple integral, I get:
$int_0^{2pi}int_0^1int_{-sqrt{3}}^sqrt{3}rdzdrdtheta$
However, Slader has a different answer where they didn't plug in $sqrt{3}$ into the bounds. Why does plugging in the value for $r^2$ make the calculation wrong? Isn't $r^2$ always $1$?
calculus integration multivariable-calculus volume multiple-integral
asked 1 hour ago
Hugh N.Hugh N.
404
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add a comment |
$begingroup$
I'm currently working on this question:
Find the volume of the solid that lies within both the cylinder $x^2+y^2=1$ and the sphere $x^2+y^2+z^2=4$.
I decided to use polar coordinates so that the cylinder equation becomes $r^2=1$ and the sphere becomes $r^2+z^2=4$.
Solving for $z$, I get the inequality $-sqrt{4-r^2}leq zleq sqrt{4-r^2}$. Since I know what $r^2$ is, I plug that in to get the inequality where $z$ is between $-sqrt{3}$ and $sqrt{3}$. Combining that to make a triple integral, I get:
$int_0^{2pi}int_0^1int_{-sqrt{3}}^sqrt{3}rdzdrdtheta$
However, Slader has a different answer where they didn't plug in $sqrt{3}$ into the bounds. Why does plugging in the value for $r^2$ make the calculation wrong? Isn't $r^2$ always $1$?
calculus integration multivariable-calculus volume multiple-integral
asked 1 hour ago
Hugh N.Hugh N.
404
$endgroup$
add a comment |
$begingroup$
I'm currently working on this question:
Find the volume of the solid that lies within both the cylinder $x^2+y^2=1$ and the sphere $x^2+y^2+z^2=4$.
I decided to use polar coordinates so that the cylinder equation becomes $r^2=1$ and the sphere becomes $r^2+z^2=4$.
Solving for $z$, I get the inequality $-sqrt{4-r^2}leq zleq sqrt{4-r^2}$. Since I know what $r^2$ is, I plug that in to get the inequality where $z$ is between $-sqrt{3}$ and $sqrt{3}$. Combining that to make a triple integral, I get:
$int_0^{2pi}int_0^1int_{-sqrt{3}}^sqrt{3}rdzdrdtheta$
However, Slader has a different answer where they didn't plug in $sqrt{3}$ into the bounds. Why does plugging in the value for $r^2$ make the calculation wrong? Isn't $r^2$ always $1$?
calculus integration multivariable-calculus volume multiple-integral
asked 1 hour ago
Hugh N.Hugh N.
404
$endgroup$
I'm currently working on this question:
Find the volume of the solid that lies within both the cylinder $x^2+y^2=1$ and the sphere $x^2+y^2+z^2=4$.
I decided to use polar coordinates so that the cylinder equation becomes $r^2=1$ and the sphere becomes $r^2+z^2=4$.
Solving for $z$, I get the inequality $-sqrt{4-r^2}leq zleq sqrt{4-r^2}$. Since I know what $r^2$ is, I plug that in to get the inequality where $z$ is between $-sqrt{3}$ and $sqrt{3}$. Combining that to make a triple integral, I get:
$int_0^{2pi}int_0^1int_{-sqrt{3}}^sqrt{3}rdzdrdtheta$
However, Slader has a different answer where they didn't plug in $sqrt{3}$ into the bounds. Why does plugging in the value for $r^2$ make the calculation wrong? Isn't $r^2$ always $1$?
calculus integration multivariable-calculus volume multiple-integral
calculus integration multivariable-calculus volume multiple-integral
asked 1 hour ago
Hugh N.Hugh N.
404
asked 1 hour ago
Hugh N.Hugh N.
404
asked 1 hour ago
Hugh N.Hugh N.
404
asked 1 hour ago
Hugh N.Hugh N.
404
asked 1 hour ago
Hugh N.Hugh N.
404
404
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2 Answers
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“Isn’t $r^2$ always 1?”
Yes, on the surface of the cylinder.
But you aren’t integrating on the surface of the cylinder. You’re integrating over a three-dimensional region, namely the one bounded by both the cylinder and the sphere. This region is described in cylindrical coordinates as the set with
$$rin[0,1]$$
$$thetain[0,2pi)$$
$$zinleft[-sqrt{4-r^2}, sqrt{4-r^2}right]$$
answered 1 hour ago
symplectomorphicsymplectomorphic
12.4k22039
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$begingroup$
No, $r$ is not always $1$. We're in the integral; $r$ is a variable we're going to integrate over later, and it doesn't have any fixed value.
The cylinder equation $r=1$ represents part of the boundary. It's certainly not true for every point inside the solid.
answered 1 hour ago
jmerryjmerry
8,7581123
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2 Answers
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2 Answers
2
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$begingroup$
“Isn’t $r^2$ always 1?”
Yes, on the surface of the cylinder.
But you aren’t integrating on the surface of the cylinder. You’re integrating over a three-dimensional region, namely the one bounded by both the cylinder and the sphere. This region is described in cylindrical coordinates as the set with
$$rin[0,1]$$
$$thetain[0,2pi)$$
$$zinleft[-sqrt{4-r^2}, sqrt{4-r^2}right]$$
answered 1 hour ago
symplectomorphicsymplectomorphic
12.4k22039
$endgroup$
add a comment |
$begingroup$
“Isn’t $r^2$ always 1?”
Yes, on the surface of the cylinder.
But you aren’t integrating on the surface of the cylinder. You’re integrating over a three-dimensional region, namely the one bounded by both the cylinder and the sphere. This region is described in cylindrical coordinates as the set with
$$rin[0,1]$$
$$thetain[0,2pi)$$
$$zinleft[-sqrt{4-r^2}, sqrt{4-r^2}right]$$
answered 1 hour ago
symplectomorphicsymplectomorphic
12.4k22039
$endgroup$
add a comment |
$begingroup$
“Isn’t $r^2$ always 1?”
Yes, on the surface of the cylinder.
But you aren’t integrating on the surface of the cylinder. You’re integrating over a three-dimensional region, namely the one bounded by both the cylinder and the sphere. This region is described in cylindrical coordinates as the set with
$$rin[0,1]$$
$$thetain[0,2pi)$$
$$zinleft[-sqrt{4-r^2}, sqrt{4-r^2}right]$$
answered 1 hour ago
symplectomorphicsymplectomorphic
12.4k22039
$endgroup$
“Isn’t $r^2$ always 1?”
Yes, on the surface of the cylinder.
But you aren’t integrating on the surface of the cylinder. You’re integrating over a three-dimensional region, namely the one bounded by both the cylinder and the sphere. This region is described in cylindrical coordinates as the set with
$$rin[0,1]$$
$$thetain[0,2pi)$$
$$zinleft[-sqrt{4-r^2}, sqrt{4-r^2}right]$$
answered 1 hour ago
symplectomorphicsymplectomorphic
12.4k22039
answered 1 hour ago
symplectomorphicsymplectomorphic
12.4k22039
answered 1 hour ago
symplectomorphicsymplectomorphic
12.4k22039
answered 1 hour ago
symplectomorphicsymplectomorphic
12.4k22039
12.4k22039
add a comment |
add a comment |
$begingroup$
No, $r$ is not always $1$. We're in the integral; $r$ is a variable we're going to integrate over later, and it doesn't have any fixed value.
The cylinder equation $r=1$ represents part of the boundary. It's certainly not true for every point inside the solid.
answered 1 hour ago
jmerryjmerry
8,7581123
$endgroup$
add a comment |
$begingroup$
No, $r$ is not always $1$. We're in the integral; $r$ is a variable we're going to integrate over later, and it doesn't have any fixed value.
The cylinder equation $r=1$ represents part of the boundary. It's certainly not true for every point inside the solid.
answered 1 hour ago
jmerryjmerry
8,7581123
$endgroup$
add a comment |
$begingroup$
No, $r$ is not always $1$. We're in the integral; $r$ is a variable we're going to integrate over later, and it doesn't have any fixed value.
The cylinder equation $r=1$ represents part of the boundary. It's certainly not true for every point inside the solid.
answered 1 hour ago
jmerryjmerry
8,7581123
$endgroup$
No, $r$ is not always $1$. We're in the integral; $r$ is a variable we're going to integrate over later, and it doesn't have any fixed value.
The cylinder equation $r=1$ represents part of the boundary. It's certainly not true for every point inside the solid.
answered 1 hour ago
jmerryjmerry
8,7581123
answered 1 hour ago
jmerryjmerry
8,7581123
answered 1 hour ago
jmerryjmerry
8,7581123
answered 1 hour ago
jmerryjmerry
8,7581123
8,7581123
add a comment |
add a comment |
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