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The limits of integration for a paraboloid and plane.
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Consider the volume, $V$, enclosed by the paraboloid
$$z=x^2+y^2$$
and the plane $2x+z=3$. Determine the limits for the enclosed volume. Do not attempt to evaluate the integral.
Hello, I’m confused on how to get limits for this integral, I drew it and I think $z$ limits are from the paraboloid equation to $z=3-2x$, but after that I’m stumped.
Thanks!
integration multivariable-calculus
edited Aug 2 at 20:21
Robert Howard
1,263620
asked Aug 2 at 19:27
Brad Scott
62
add a comment |Â
up vote
1
down vote
favorite
Consider the volume, $V$, enclosed by the paraboloid
$$z=x^2+y^2$$
and the plane $2x+z=3$. Determine the limits for the enclosed volume. Do not attempt to evaluate the integral.
Hello, I’m confused on how to get limits for this integral, I drew it and I think $z$ limits are from the paraboloid equation to $z=3-2x$, but after that I’m stumped.
Thanks!
integration multivariable-calculus
edited Aug 2 at 20:21
Robert Howard
1,263620
asked Aug 2 at 19:27
Brad Scott
62
I wonder what they are looking for. The "volume $V$" (I would have written "body $K$" or something) is really given as the volume between the paraboloid and the plane.
– mickep
Aug 2 at 19:34
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Consider the volume, $V$, enclosed by the paraboloid
$$z=x^2+y^2$$
and the plane $2x+z=3$. Determine the limits for the enclosed volume. Do not attempt to evaluate the integral.
Hello, I’m confused on how to get limits for this integral, I drew it and I think $z$ limits are from the paraboloid equation to $z=3-2x$, but after that I’m stumped.
Thanks!
integration multivariable-calculus
edited Aug 2 at 20:21
Robert Howard
1,263620
asked Aug 2 at 19:27
Brad Scott
62
Consider the volume, $V$, enclosed by the paraboloid
$$z=x^2+y^2$$
and the plane $2x+z=3$. Determine the limits for the enclosed volume. Do not attempt to evaluate the integral.
Hello, I’m confused on how to get limits for this integral, I drew it and I think $z$ limits are from the paraboloid equation to $z=3-2x$, but after that I’m stumped.
Thanks!
integration multivariable-calculus
edited Aug 2 at 20:21
Robert Howard
1,263620
asked Aug 2 at 19:27
Brad Scott
62
edited Aug 2 at 20:21
Robert Howard
1,263620
edited Aug 2 at 20:21
Robert Howard
1,263620
edited Aug 2 at 20:21
Robert Howard
1,263620
1,263620
asked Aug 2 at 19:27
Brad Scott
62
asked Aug 2 at 19:27
Brad Scott
62
asked Aug 2 at 19:27
Brad Scott
62
62
I wonder what they are looking for. The "volume $V$" (I would have written "body $K$" or something) is really given as the volume between the paraboloid and the plane.
– mickep
Aug 2 at 19:34
add a comment |Â
I wonder what they are looking for. The "volume $V$" (I would have written "body $K$" or something) is really given as the volume between the paraboloid and the plane.
– mickep
Aug 2 at 19:34
I wonder what they are looking for. The "volume $V$" (I would have written "body $K$" or something) is really given as the volume between the paraboloid and the plane.
– mickep
Aug 2 at 19:34
I wonder what they are looking for. The "volume $V$" (I would have written "body $K$" or something) is really given as the volume between the paraboloid and the plane.
– mickep
Aug 2 at 19:34
add a comment |Â
2 Answers
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Well, first we can find the $x$ and $y$ coordinates that we'll have to integrate across by solving for the intersection of the plane and the paraboloid. To do this, we rearrange our equations as $z=3-2x$ and $z=x^2+y^2$ and set them equal to get
$$3-2x=x^2+y^2 implies 4=(x+1)^2+y^2$$
So our region is a circle of radius $2$ around the point $(x,y)=(-1,0)$. Now we have our limits by rearranging our circle into $x=-1 pmsqrt4-y^2$. The $z$ values that we will integrate across will simply be from the paraboloid to the plane so we can write
$$int_-2^2 int_-1-sqrt4-y^2^-1+sqrt4-y^2int_x^2+y^2^3-2x dz dx dy$$
or
$$int_-2^2 int_-1-sqrt4-y^2^-1+sqrt4-y^2(3-2x) - (x^2+y^2) dx dy$$
answered Aug 3 at 3:34
Isaac Browne
3,7112928
add a comment |Â
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The two surfaces intersect when $$z=x^2+y^2=3-2x Rightarrow x^2+2x+y^2 = 3$$
Now $$x^2+2x+y^2 = 3 Rightarrow left(x+1right)^2+y^2=4$$
Therefore in cartesian coordinates:
$$V= intintint dzdydx=int_x=-3^x=1int_y=-sqrt4-(x+1)^2^y=sqrt4-(x+1)^2int_z=x^2+y^2^z=3-2x1dzdydx$$
In Cylindrical coordinates, starting from:
$$V= intintint_z=x^2+y^2^z=3-2x1dzdA=intintleft[4-left(x+1right)^2-y^2right]dA$$
Then using $r=x^2+y^2$, $x=rcostheta$ and $y=rsintheta$:
$$V=int_theta=0^theta=2piint_r=0^r=2left[ 4-left(rcostheta+1right)^2-left( rsintheta right)^2 right]rdrdtheta$$
answered Aug 3 at 3:50
Winter Soldier
17828
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Well, first we can find the $x$ and $y$ coordinates that we'll have to integrate across by solving for the intersection of the plane and the paraboloid. To do this, we rearrange our equations as $z=3-2x$ and $z=x^2+y^2$ and set them equal to get
$$3-2x=x^2+y^2 implies 4=(x+1)^2+y^2$$
So our region is a circle of radius $2$ around the point $(x,y)=(-1,0)$. Now we have our limits by rearranging our circle into $x=-1 pmsqrt4-y^2$. The $z$ values that we will integrate across will simply be from the paraboloid to the plane so we can write
$$int_-2^2 int_-1-sqrt4-y^2^-1+sqrt4-y^2int_x^2+y^2^3-2x dz dx dy$$
or
$$int_-2^2 int_-1-sqrt4-y^2^-1+sqrt4-y^2(3-2x) - (x^2+y^2) dx dy$$
answered Aug 3 at 3:34
Isaac Browne
3,7112928
add a comment |Â
up vote
0
down vote
Well, first we can find the $x$ and $y$ coordinates that we'll have to integrate across by solving for the intersection of the plane and the paraboloid. To do this, we rearrange our equations as $z=3-2x$ and $z=x^2+y^2$ and set them equal to get
$$3-2x=x^2+y^2 implies 4=(x+1)^2+y^2$$
So our region is a circle of radius $2$ around the point $(x,y)=(-1,0)$. Now we have our limits by rearranging our circle into $x=-1 pmsqrt4-y^2$. The $z$ values that we will integrate across will simply be from the paraboloid to the plane so we can write
$$int_-2^2 int_-1-sqrt4-y^2^-1+sqrt4-y^2int_x^2+y^2^3-2x dz dx dy$$
or
$$int_-2^2 int_-1-sqrt4-y^2^-1+sqrt4-y^2(3-2x) - (x^2+y^2) dx dy$$
answered Aug 3 at 3:34
Isaac Browne
3,7112928
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Well, first we can find the $x$ and $y$ coordinates that we'll have to integrate across by solving for the intersection of the plane and the paraboloid. To do this, we rearrange our equations as $z=3-2x$ and $z=x^2+y^2$ and set them equal to get
$$3-2x=x^2+y^2 implies 4=(x+1)^2+y^2$$
So our region is a circle of radius $2$ around the point $(x,y)=(-1,0)$. Now we have our limits by rearranging our circle into $x=-1 pmsqrt4-y^2$. The $z$ values that we will integrate across will simply be from the paraboloid to the plane so we can write
$$int_-2^2 int_-1-sqrt4-y^2^-1+sqrt4-y^2int_x^2+y^2^3-2x dz dx dy$$
or
$$int_-2^2 int_-1-sqrt4-y^2^-1+sqrt4-y^2(3-2x) - (x^2+y^2) dx dy$$
answered Aug 3 at 3:34
Isaac Browne
3,7112928
Well, first we can find the $x$ and $y$ coordinates that we'll have to integrate across by solving for the intersection of the plane and the paraboloid. To do this, we rearrange our equations as $z=3-2x$ and $z=x^2+y^2$ and set them equal to get
$$3-2x=x^2+y^2 implies 4=(x+1)^2+y^2$$
So our region is a circle of radius $2$ around the point $(x,y)=(-1,0)$. Now we have our limits by rearranging our circle into $x=-1 pmsqrt4-y^2$. The $z$ values that we will integrate across will simply be from the paraboloid to the plane so we can write
$$int_-2^2 int_-1-sqrt4-y^2^-1+sqrt4-y^2int_x^2+y^2^3-2x dz dx dy$$
or
$$int_-2^2 int_-1-sqrt4-y^2^-1+sqrt4-y^2(3-2x) - (x^2+y^2) dx dy$$
answered Aug 3 at 3:34
Isaac Browne
3,7112928
answered Aug 3 at 3:34
Isaac Browne
3,7112928
answered Aug 3 at 3:34
Isaac Browne
3,7112928
answered Aug 3 at 3:34
Isaac Browne
3,7112928
3,7112928
add a comment |Â
add a comment |Â
up vote
0
down vote
The two surfaces intersect when $$z=x^2+y^2=3-2x Rightarrow x^2+2x+y^2 = 3$$
Now $$x^2+2x+y^2 = 3 Rightarrow left(x+1right)^2+y^2=4$$
Therefore in cartesian coordinates:
$$V= intintint dzdydx=int_x=-3^x=1int_y=-sqrt4-(x+1)^2^y=sqrt4-(x+1)^2int_z=x^2+y^2^z=3-2x1dzdydx$$
In Cylindrical coordinates, starting from:
$$V= intintint_z=x^2+y^2^z=3-2x1dzdA=intintleft[4-left(x+1right)^2-y^2right]dA$$
Then using $r=x^2+y^2$, $x=rcostheta$ and $y=rsintheta$:
$$V=int_theta=0^theta=2piint_r=0^r=2left[ 4-left(rcostheta+1right)^2-left( rsintheta right)^2 right]rdrdtheta$$
answered Aug 3 at 3:50
Winter Soldier
17828
add a comment |Â
up vote
0
down vote
The two surfaces intersect when $$z=x^2+y^2=3-2x Rightarrow x^2+2x+y^2 = 3$$
Now $$x^2+2x+y^2 = 3 Rightarrow left(x+1right)^2+y^2=4$$
Therefore in cartesian coordinates:
$$V= intintint dzdydx=int_x=-3^x=1int_y=-sqrt4-(x+1)^2^y=sqrt4-(x+1)^2int_z=x^2+y^2^z=3-2x1dzdydx$$
In Cylindrical coordinates, starting from:
$$V= intintint_z=x^2+y^2^z=3-2x1dzdA=intintleft[4-left(x+1right)^2-y^2right]dA$$
Then using $r=x^2+y^2$, $x=rcostheta$ and $y=rsintheta$:
$$V=int_theta=0^theta=2piint_r=0^r=2left[ 4-left(rcostheta+1right)^2-left( rsintheta right)^2 right]rdrdtheta$$
answered Aug 3 at 3:50
Winter Soldier
17828
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The two surfaces intersect when $$z=x^2+y^2=3-2x Rightarrow x^2+2x+y^2 = 3$$
Now $$x^2+2x+y^2 = 3 Rightarrow left(x+1right)^2+y^2=4$$
Therefore in cartesian coordinates:
$$V= intintint dzdydx=int_x=-3^x=1int_y=-sqrt4-(x+1)^2^y=sqrt4-(x+1)^2int_z=x^2+y^2^z=3-2x1dzdydx$$
In Cylindrical coordinates, starting from:
$$V= intintint_z=x^2+y^2^z=3-2x1dzdA=intintleft[4-left(x+1right)^2-y^2right]dA$$
Then using $r=x^2+y^2$, $x=rcostheta$ and $y=rsintheta$:
$$V=int_theta=0^theta=2piint_r=0^r=2left[ 4-left(rcostheta+1right)^2-left( rsintheta right)^2 right]rdrdtheta$$
answered Aug 3 at 3:50
Winter Soldier
17828
The two surfaces intersect when $$z=x^2+y^2=3-2x Rightarrow x^2+2x+y^2 = 3$$
Now $$x^2+2x+y^2 = 3 Rightarrow left(x+1right)^2+y^2=4$$
Therefore in cartesian coordinates:
$$V= intintint dzdydx=int_x=-3^x=1int_y=-sqrt4-(x+1)^2^y=sqrt4-(x+1)^2int_z=x^2+y^2^z=3-2x1dzdydx$$
In Cylindrical coordinates, starting from:
$$V= intintint_z=x^2+y^2^z=3-2x1dzdA=intintleft[4-left(x+1right)^2-y^2right]dA$$
Then using $r=x^2+y^2$, $x=rcostheta$ and $y=rsintheta$:
$$V=int_theta=0^theta=2piint_r=0^r=2left[ 4-left(rcostheta+1right)^2-left( rsintheta right)^2 right]rdrdtheta$$
answered Aug 3 at 3:50
Winter Soldier
17828
answered Aug 3 at 3:50
Winter Soldier
17828
answered Aug 3 at 3:50
Winter Soldier
17828
answered Aug 3 at 3:50
Winter Soldier
17828
17828
add a comment |Â
add a comment |Â
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I wonder what they are looking for. The "volume $V$" (I would have written "body $K$" or something) is really given as the volume between the paraboloid and the plane.
– mickep
Aug 2 at 19:34