Find the volume bounded by $C$ and the planes ( integrals ) ( question 9 )












1












$begingroup$



Let $C$ be the cylinder $big{(x,y,z):x^2+y^2leq 1big}$. Find the volume bounded by $C$ above the plane $z=0$ and below the plane $x+z=1$.




enter image description here



I have wondered how to solve this , here is my attempt :




$$
int_{x=-1}^{x=1}int_{z=0}^{z=1-x}int_{y=-sqrt{1-x^2}}^{y=sqrt{1-x^2}} 1{ dy}{ dz}{ dx} = pi$$




i dont know how to solve these kind of questions do i have to sketch the domain ?
or to solve it using algebric inequality






calculus integration multivariable-calculus volume







share|cite|improve this question











$endgroup$












  • $begingroup$
    can somebody help please
    $endgroup$
    – Mather
    Dec 15 '18 at 13:43










  • $begingroup$
    how do i find the boundaries ? i tried to sketch but i cant decidee the boundaries just by looking on the sketch right ?
    $endgroup$
    – Mather
    Dec 15 '18 at 13:56










  • $begingroup$
    i used both the poolar and the way i wrote up there and i got pi is it the right answer ?
    $endgroup$
    – Mather
    Dec 15 '18 at 14:13
















1












$begingroup$



Let $C$ be the cylinder $big{(x,y,z):x^2+y^2leq 1big}$. Find the volume bounded by $C$ above the plane $z=0$ and below the plane $x+z=1$.




enter image description here



I have wondered how to solve this , here is my attempt :




$$
int_{x=-1}^{x=1}int_{z=0}^{z=1-x}int_{y=-sqrt{1-x^2}}^{y=sqrt{1-x^2}} 1{ dy}{ dz}{ dx} = pi$$




i dont know how to solve these kind of questions do i have to sketch the domain ?
or to solve it using algebric inequality






calculus integration multivariable-calculus volume







share|cite|improve this question











$endgroup$












  • $begingroup$
    can somebody help please
    $endgroup$
    – Mather
    Dec 15 '18 at 13:43










  • $begingroup$
    how do i find the boundaries ? i tried to sketch but i cant decidee the boundaries just by looking on the sketch right ?
    $endgroup$
    – Mather
    Dec 15 '18 at 13:56










  • $begingroup$
    i used both the poolar and the way i wrote up there and i got pi is it the right answer ?
    $endgroup$
    – Mather
    Dec 15 '18 at 14:13














1












1








1





$begingroup$



Let $C$ be the cylinder $big{(x,y,z):x^2+y^2leq 1big}$. Find the volume bounded by $C$ above the plane $z=0$ and below the plane $x+z=1$.




enter image description here



I have wondered how to solve this , here is my attempt :




$$
int_{x=-1}^{x=1}int_{z=0}^{z=1-x}int_{y=-sqrt{1-x^2}}^{y=sqrt{1-x^2}} 1{ dy}{ dz}{ dx} = pi$$




i dont know how to solve these kind of questions do i have to sketch the domain ?
or to solve it using algebric inequality






calculus integration multivariable-calculus volume







share|cite|improve this question











$endgroup$





Let $C$ be the cylinder $big{(x,y,z):x^2+y^2leq 1big}$. Find the volume bounded by $C$ above the plane $z=0$ and below the plane $x+z=1$.




enter image description here



I have wondered how to solve this , here is my attempt :




$$
int_{x=-1}^{x=1}int_{z=0}^{z=1-x}int_{y=-sqrt{1-x^2}}^{y=sqrt{1-x^2}} 1{ dy}{ dz}{ dx} = pi$$




i dont know how to solve these kind of questions do i have to sketch the domain ?
or to solve it using algebric inequality






calculus integration multivariable-calculus volume




calculus integration multivariable-calculus volume






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 15 '18 at 17:48







user593746

















asked Dec 15 '18 at 11:15









Mather Mather

3317




3317












  • $begingroup$
    can somebody help please
    $endgroup$
    – Mather
    Dec 15 '18 at 13:43










  • $begingroup$
    how do i find the boundaries ? i tried to sketch but i cant decidee the boundaries just by looking on the sketch right ?
    $endgroup$
    – Mather
    Dec 15 '18 at 13:56










  • $begingroup$
    i used both the poolar and the way i wrote up there and i got pi is it the right answer ?
    $endgroup$
    – Mather
    Dec 15 '18 at 14:13


















  • $begingroup$
    can somebody help please
    $endgroup$
    – Mather
    Dec 15 '18 at 13:43










  • $begingroup$
    how do i find the boundaries ? i tried to sketch but i cant decidee the boundaries just by looking on the sketch right ?
    $endgroup$
    – Mather
    Dec 15 '18 at 13:56










  • $begingroup$
    i used both the poolar and the way i wrote up there and i got pi is it the right answer ?
    $endgroup$
    – Mather
    Dec 15 '18 at 14:13
















$begingroup$
can somebody help please
$endgroup$
– Mather
Dec 15 '18 at 13:43




$begingroup$
can somebody help please
$endgroup$
– Mather
Dec 15 '18 at 13:43












$begingroup$
how do i find the boundaries ? i tried to sketch but i cant decidee the boundaries just by looking on the sketch right ?
$endgroup$
– Mather
Dec 15 '18 at 13:56




$begingroup$
how do i find the boundaries ? i tried to sketch but i cant decidee the boundaries just by looking on the sketch right ?
$endgroup$
– Mather
Dec 15 '18 at 13:56












$begingroup$
i used both the poolar and the way i wrote up there and i got pi is it the right answer ?
$endgroup$
– Mather
Dec 15 '18 at 14:13




$begingroup$
i used both the poolar and the way i wrote up there and i got pi is it the right answer ?
$endgroup$
– Mather
Dec 15 '18 at 14:13










1 Answer
1






active

oldest

votes


















1












$begingroup$

Let $Z$ be the finite cylinder $$big{(x,y,z)inBbb R^3:x^2+y^2leq 1wedge 0leq zleq 2}.$$ Let $B$ denote the required region which is $$big{(x,y,z)inBbb{R}^3:x^2+y^2leq 1wedge 0leq zleq 1-xbig}.$$ Let $B'$ denote the closure of $Zsetminus B$. Hence, $$B'= big{(x,y,z)inBbb{R}^3:x^2+y^2leq 1wedge 1-xleq zleq 2}.$$ Then, the following isometry $f:mathbb{R}^3to mathbb{R}^3$ maps $B$ to $B'$:
$$f(x,y,z)=(-x,y,2-z).$$
Therefore, $operatorname{vol}B=operatorname{vol}B'$. Since $Bcap B'$ is a subset of the plane given by $x+z=1$, $operatorname{vol}(Bcap B')=0$. Because $Z=Bcup B'$, we obtain
$$operatorname{vol}Z=operatorname{vol}B+operatorname{vol}B'-operatorname{vol}(Bcap B')=2operatorname{vol}B.$$
As $operatorname{vol}Z=picdot 1^2cdot 2=2pi$, we get
$$operatorname{vol}B=pi.$$
This confirms that your solution is correct.



Your solution isn't bad either. Disengaging $y$ and $z$ from your integral, the volume of $B$ is
$$2int_{-1}^1 (1-x)sqrt{1-x^2} dx=left.left(frac{(-2x^2+3x+2)sqrt{1-x^2}+3arcsin x}{3}right)right|_{-1}^1=pi.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    thank you @Zvi, can you verify one last question here please if you could , math.stackexchange.com/q/3041581/615298
    $endgroup$
    – Mather
    Dec 15 '18 at 18:38






  • 1




    $begingroup$
    Unfortunately, I am leaving my office for the day and don't have time for the evening. If the thread is still unanswered tomorrow, I will try to answer it.
    $endgroup$
    – user593746
    Dec 15 '18 at 19:32











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Let $Z$ be the finite cylinder $$big{(x,y,z)inBbb R^3:x^2+y^2leq 1wedge 0leq zleq 2}.$$ Let $B$ denote the required region which is $$big{(x,y,z)inBbb{R}^3:x^2+y^2leq 1wedge 0leq zleq 1-xbig}.$$ Let $B'$ denote the closure of $Zsetminus B$. Hence, $$B'= big{(x,y,z)inBbb{R}^3:x^2+y^2leq 1wedge 1-xleq zleq 2}.$$ Then, the following isometry $f:mathbb{R}^3to mathbb{R}^3$ maps $B$ to $B'$:
$$f(x,y,z)=(-x,y,2-z).$$
Therefore, $operatorname{vol}B=operatorname{vol}B'$. Since $Bcap B'$ is a subset of the plane given by $x+z=1$, $operatorname{vol}(Bcap B')=0$. Because $Z=Bcup B'$, we obtain
$$operatorname{vol}Z=operatorname{vol}B+operatorname{vol}B'-operatorname{vol}(Bcap B')=2operatorname{vol}B.$$
As $operatorname{vol}Z=picdot 1^2cdot 2=2pi$, we get
$$operatorname{vol}B=pi.$$
This confirms that your solution is correct.



Your solution isn't bad either. Disengaging $y$ and $z$ from your integral, the volume of $B$ is
$$2int_{-1}^1 (1-x)sqrt{1-x^2} dx=left.left(frac{(-2x^2+3x+2)sqrt{1-x^2}+3arcsin x}{3}right)right|_{-1}^1=pi.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    thank you @Zvi, can you verify one last question here please if you could , math.stackexchange.com/q/3041581/615298
    $endgroup$
    – Mather
    Dec 15 '18 at 18:38






  • 1




    $begingroup$
    Unfortunately, I am leaving my office for the day and don't have time for the evening. If the thread is still unanswered tomorrow, I will try to answer it.
    $endgroup$
    – user593746
    Dec 15 '18 at 19:32
















1












$begingroup$

Let $Z$ be the finite cylinder $$big{(x,y,z)inBbb R^3:x^2+y^2leq 1wedge 0leq zleq 2}.$$ Let $B$ denote the required region which is $$big{(x,y,z)inBbb{R}^3:x^2+y^2leq 1wedge 0leq zleq 1-xbig}.$$ Let $B'$ denote the closure of $Zsetminus B$. Hence, $$B'= big{(x,y,z)inBbb{R}^3:x^2+y^2leq 1wedge 1-xleq zleq 2}.$$ Then, the following isometry $f:mathbb{R}^3to mathbb{R}^3$ maps $B$ to $B'$:
$$f(x,y,z)=(-x,y,2-z).$$
Therefore, $operatorname{vol}B=operatorname{vol}B'$. Since $Bcap B'$ is a subset of the plane given by $x+z=1$, $operatorname{vol}(Bcap B')=0$. Because $Z=Bcup B'$, we obtain
$$operatorname{vol}Z=operatorname{vol}B+operatorname{vol}B'-operatorname{vol}(Bcap B')=2operatorname{vol}B.$$
As $operatorname{vol}Z=picdot 1^2cdot 2=2pi$, we get
$$operatorname{vol}B=pi.$$
This confirms that your solution is correct.



Your solution isn't bad either. Disengaging $y$ and $z$ from your integral, the volume of $B$ is
$$2int_{-1}^1 (1-x)sqrt{1-x^2} dx=left.left(frac{(-2x^2+3x+2)sqrt{1-x^2}+3arcsin x}{3}right)right|_{-1}^1=pi.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    thank you @Zvi, can you verify one last question here please if you could , math.stackexchange.com/q/3041581/615298
    $endgroup$
    – Mather
    Dec 15 '18 at 18:38






  • 1




    $begingroup$
    Unfortunately, I am leaving my office for the day and don't have time for the evening. If the thread is still unanswered tomorrow, I will try to answer it.
    $endgroup$
    – user593746
    Dec 15 '18 at 19:32














1












1








1





$begingroup$

Let $Z$ be the finite cylinder $$big{(x,y,z)inBbb R^3:x^2+y^2leq 1wedge 0leq zleq 2}.$$ Let $B$ denote the required region which is $$big{(x,y,z)inBbb{R}^3:x^2+y^2leq 1wedge 0leq zleq 1-xbig}.$$ Let $B'$ denote the closure of $Zsetminus B$. Hence, $$B'= big{(x,y,z)inBbb{R}^3:x^2+y^2leq 1wedge 1-xleq zleq 2}.$$ Then, the following isometry $f:mathbb{R}^3to mathbb{R}^3$ maps $B$ to $B'$:
$$f(x,y,z)=(-x,y,2-z).$$
Therefore, $operatorname{vol}B=operatorname{vol}B'$. Since $Bcap B'$ is a subset of the plane given by $x+z=1$, $operatorname{vol}(Bcap B')=0$. Because $Z=Bcup B'$, we obtain
$$operatorname{vol}Z=operatorname{vol}B+operatorname{vol}B'-operatorname{vol}(Bcap B')=2operatorname{vol}B.$$
As $operatorname{vol}Z=picdot 1^2cdot 2=2pi$, we get
$$operatorname{vol}B=pi.$$
This confirms that your solution is correct.



Your solution isn't bad either. Disengaging $y$ and $z$ from your integral, the volume of $B$ is
$$2int_{-1}^1 (1-x)sqrt{1-x^2} dx=left.left(frac{(-2x^2+3x+2)sqrt{1-x^2}+3arcsin x}{3}right)right|_{-1}^1=pi.$$






share|cite|improve this answer











$endgroup$



Let $Z$ be the finite cylinder $$big{(x,y,z)inBbb R^3:x^2+y^2leq 1wedge 0leq zleq 2}.$$ Let $B$ denote the required region which is $$big{(x,y,z)inBbb{R}^3:x^2+y^2leq 1wedge 0leq zleq 1-xbig}.$$ Let $B'$ denote the closure of $Zsetminus B$. Hence, $$B'= big{(x,y,z)inBbb{R}^3:x^2+y^2leq 1wedge 1-xleq zleq 2}.$$ Then, the following isometry $f:mathbb{R}^3to mathbb{R}^3$ maps $B$ to $B'$:
$$f(x,y,z)=(-x,y,2-z).$$
Therefore, $operatorname{vol}B=operatorname{vol}B'$. Since $Bcap B'$ is a subset of the plane given by $x+z=1$, $operatorname{vol}(Bcap B')=0$. Because $Z=Bcup B'$, we obtain
$$operatorname{vol}Z=operatorname{vol}B+operatorname{vol}B'-operatorname{vol}(Bcap B')=2operatorname{vol}B.$$
As $operatorname{vol}Z=picdot 1^2cdot 2=2pi$, we get
$$operatorname{vol}B=pi.$$
This confirms that your solution is correct.



Your solution isn't bad either. Disengaging $y$ and $z$ from your integral, the volume of $B$ is
$$2int_{-1}^1 (1-x)sqrt{1-x^2} dx=left.left(frac{(-2x^2+3x+2)sqrt{1-x^2}+3arcsin x}{3}right)right|_{-1}^1=pi.$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 15 '18 at 18:08

























answered Dec 15 '18 at 18:02







user593746



















  • $begingroup$
    thank you @Zvi, can you verify one last question here please if you could , math.stackexchange.com/q/3041581/615298
    $endgroup$
    – Mather
    Dec 15 '18 at 18:38






  • 1




    $begingroup$
    Unfortunately, I am leaving my office for the day and don't have time for the evening. If the thread is still unanswered tomorrow, I will try to answer it.
    $endgroup$
    – user593746
    Dec 15 '18 at 19:32


















  • $begingroup$
    thank you @Zvi, can you verify one last question here please if you could , math.stackexchange.com/q/3041581/615298
    $endgroup$
    – Mather
    Dec 15 '18 at 18:38






  • 1




    $begingroup$
    Unfortunately, I am leaving my office for the day and don't have time for the evening. If the thread is still unanswered tomorrow, I will try to answer it.
    $endgroup$
    – user593746
    Dec 15 '18 at 19:32
















$begingroup$
thank you @Zvi, can you verify one last question here please if you could , math.stackexchange.com/q/3041581/615298
$endgroup$
– Mather
Dec 15 '18 at 18:38




$begingroup$
thank you @Zvi, can you verify one last question here please if you could , math.stackexchange.com/q/3041581/615298
$endgroup$
– Mather
Dec 15 '18 at 18:38




1




1




$begingroup$
Unfortunately, I am leaving my office for the day and don't have time for the evening. If the thread is still unanswered tomorrow, I will try to answer it.
$endgroup$
– user593746
Dec 15 '18 at 19:32




$begingroup$
Unfortunately, I am leaving my office for the day and don't have time for the evening. If the thread is still unanswered tomorrow, I will try to answer it.
$endgroup$
– user593746
Dec 15 '18 at 19:32


















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