Creating an integral from 2 functions The 2019 Stack Overflow Developer Survey Results Are InSketch the region $R$ and evaluate the double integral $iint 2y;mathrm dA$Taking the line integral of a region with holes with Green's TheoremCalculating Flux with Gauss's TheoremFinding the Limits of the Triple Integral (Spherical Coordinates)Volume integral over a wedge shaped cylindrical regionComposing an iterated double integral given the equation of three lines which form a triangle.Integral $int (t^2 - 1)^a cdot t^b cdot log(t),dt$Is this really a double integral problem?Area under curve: integration
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Creating an integral from 2 functions
The 2019 Stack Overflow Developer Survey Results Are InSketch the region $R$ and evaluate the double integral $iint 2y;mathrm dA$Taking the line integral of a region with holes with Green's TheoremCalculating Flux with Gauss's TheoremFinding the Limits of the Triple Integral (Spherical Coordinates)Volume integral over a wedge shaped cylindrical regionComposing an iterated double integral given the equation of three lines which form a triangle.Integral $int (t^2 - 1)^a cdot t^b cdot log(t),dt$Is this really a double integral problem?Area under curve: integration
$begingroup$
Let $R$ be the region in $mathbbR^2$ below the line $y = x + 2$ and above the parabola $y = x^2$. Check the integral of these $2$ functions in terms of $dxcdot dy$ and then $dycdot dx$
'
I am having an issue figuring out what the integrals will range from. I have:
$$G =(y,x) : -1 < x < 2 text and x^2 < y < (x + 2)to dy.dx$$
$$H =(x,y) : 0 < y < 4 text and y -2 < x < sqrty to dx.dy$$
However when I create the integrals in terms of $dxcdot dy$ and $dycdot dx$ they differ? Any help please, did i get the range of the $G$ and $H$ wrong? Its a parabola cut with a line
integration
edited Mar 30 at 19:12
Siong Thye Goh
104k1468120
asked Mar 30 at 17:11
Shaun WeinbergShaun Weinberg
144
$endgroup$
add a comment |
$begingroup$
Let $R$ be the region in $mathbbR^2$ below the line $y = x + 2$ and above the parabola $y = x^2$. Check the integral of these $2$ functions in terms of $dxcdot dy$ and then $dycdot dx$
'
I am having an issue figuring out what the integrals will range from. I have:
$$G =(y,x) : -1 < x < 2 text and x^2 < y < (x + 2)to dy.dx$$
$$H =(x,y) : 0 < y < 4 text and y -2 < x < sqrty to dx.dy$$
However when I create the integrals in terms of $dxcdot dy$ and $dycdot dx$ they differ? Any help please, did i get the range of the $G$ and $H$ wrong? Its a parabola cut with a line
integration
edited Mar 30 at 19:12
Siong Thye Goh
104k1468120
asked Mar 30 at 17:11
Shaun WeinbergShaun Weinberg
144
$endgroup$
add a comment |
$begingroup$
Let $R$ be the region in $mathbbR^2$ below the line $y = x + 2$ and above the parabola $y = x^2$. Check the integral of these $2$ functions in terms of $dxcdot dy$ and then $dycdot dx$
'
I am having an issue figuring out what the integrals will range from. I have:
$$G =(y,x) : -1 < x < 2 text and x^2 < y < (x + 2)to dy.dx$$
$$H =(x,y) : 0 < y < 4 text and y -2 < x < sqrty to dx.dy$$
However when I create the integrals in terms of $dxcdot dy$ and $dycdot dx$ they differ? Any help please, did i get the range of the $G$ and $H$ wrong? Its a parabola cut with a line
integration
edited Mar 30 at 19:12
Siong Thye Goh
104k1468120
asked Mar 30 at 17:11
Shaun WeinbergShaun Weinberg
144
$endgroup$
Let $R$ be the region in $mathbbR^2$ below the line $y = x + 2$ and above the parabola $y = x^2$. Check the integral of these $2$ functions in terms of $dxcdot dy$ and then $dycdot dx$
'
I am having an issue figuring out what the integrals will range from. I have:
$$G =(y,x) : -1 < x < 2 text and x^2 < y < (x + 2)to dy.dx$$
$$H =(x,y) : 0 < y < 4 text and y -2 < x < sqrty to dx.dy$$
However when I create the integrals in terms of $dxcdot dy$ and $dycdot dx$ they differ? Any help please, did i get the range of the $G$ and $H$ wrong? Its a parabola cut with a line
integration
integration
edited Mar 30 at 19:12
Siong Thye Goh
104k1468120
asked Mar 30 at 17:11
Shaun WeinbergShaun Weinberg
144
edited Mar 30 at 19:12
Siong Thye Goh
104k1468120
asked Mar 30 at 17:11
Shaun WeinbergShaun Weinberg
144
edited Mar 30 at 19:12
Siong Thye Goh
104k1468120
edited Mar 30 at 19:12
Siong Thye Goh
104k1468120
edited Mar 30 at 19:12
Siong Thye Goh
104k1468120
104k1468120
asked Mar 30 at 17:11
Shaun WeinbergShaun Weinberg
144
asked Mar 30 at 17:11
Shaun WeinbergShaun Weinberg
144
asked Mar 30 at 17:11
Shaun WeinbergShaun Weinberg
144
144
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
For $dydx$ the integral over $1$ is given by
$$int_-1^2int_x^2^x+2dydx=int_-1^2-x^2+x+2dx=[frac13x^3+frac12x^2+2x]_-1^2=frac92$$
with the region that you correctly evaluated. But for the second integral we need to split the region into two parts - for $0le yle1$ we have that $-sqrtyle xlesqrty$ and when $1le yle4$ we have $y-2le xlesqrty$. So the integral over the function $1$ is
$$int_0^1int_-sqrty^sqrtydxdy+int_1^4int_y-2^sqrtydxdy=int_0^12sqrty,dy+int_1^4sqrty-y+2,dy$$
$$=[frac43y^frac32]_0^1+[frac23y^frac32-frac12y^2+2y]_1^4=frac92$$
So the two regions are now equal.
answered Mar 30 at 19:12
Peter ForemanPeter Foreman
6,9401318
$endgroup$
$begingroup$
Thank you this helps loads. My text book does not describe splitting them up and i did not think of it. Much appreciated!
$endgroup$
– Shaun Weinberg
Mar 31 at 10:36
add a comment |
$begingroup$
For the region $H$, the lower limit of $x$ shouldn't be $y-2$.
It should be the maximum of $-sqrty$ and $y-2$. In fact, when $0 le y le 1$, the lower limit is $-sqrty$.
That is
$$H =(x,y): 0 le y le 4 , max(-sqrty, y-2) < x < sqrty $$
That is from the picture below, the left limit of the region consists of the green color and blue color part.
answered Mar 30 at 19:09
Siong Thye GohSiong Thye Goh
104k1468120
$endgroup$
$begingroup$
I originally wrote an answer for the duplicate, so I copied it here. (+1) btw
$endgroup$
– Peter Foreman
Mar 30 at 19:13
$begingroup$
I did the same thing. ;)
$endgroup$
– Siong Thye Goh
Mar 30 at 19:14
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For $dydx$ the integral over $1$ is given by
$$int_-1^2int_x^2^x+2dydx=int_-1^2-x^2+x+2dx=[frac13x^3+frac12x^2+2x]_-1^2=frac92$$
with the region that you correctly evaluated. But for the second integral we need to split the region into two parts - for $0le yle1$ we have that $-sqrtyle xlesqrty$ and when $1le yle4$ we have $y-2le xlesqrty$. So the integral over the function $1$ is
$$int_0^1int_-sqrty^sqrtydxdy+int_1^4int_y-2^sqrtydxdy=int_0^12sqrty,dy+int_1^4sqrty-y+2,dy$$
$$=[frac43y^frac32]_0^1+[frac23y^frac32-frac12y^2+2y]_1^4=frac92$$
So the two regions are now equal.
answered Mar 30 at 19:12
Peter ForemanPeter Foreman
6,9401318
$endgroup$
$begingroup$
Thank you this helps loads. My text book does not describe splitting them up and i did not think of it. Much appreciated!
$endgroup$
– Shaun Weinberg
Mar 31 at 10:36
add a comment |
$begingroup$
For $dydx$ the integral over $1$ is given by
$$int_-1^2int_x^2^x+2dydx=int_-1^2-x^2+x+2dx=[frac13x^3+frac12x^2+2x]_-1^2=frac92$$
with the region that you correctly evaluated. But for the second integral we need to split the region into two parts - for $0le yle1$ we have that $-sqrtyle xlesqrty$ and when $1le yle4$ we have $y-2le xlesqrty$. So the integral over the function $1$ is
$$int_0^1int_-sqrty^sqrtydxdy+int_1^4int_y-2^sqrtydxdy=int_0^12sqrty,dy+int_1^4sqrty-y+2,dy$$
$$=[frac43y^frac32]_0^1+[frac23y^frac32-frac12y^2+2y]_1^4=frac92$$
So the two regions are now equal.
answered Mar 30 at 19:12
Peter ForemanPeter Foreman
6,9401318
$endgroup$
$begingroup$
Thank you this helps loads. My text book does not describe splitting them up and i did not think of it. Much appreciated!
$endgroup$
– Shaun Weinberg
Mar 31 at 10:36
add a comment |
$begingroup$
For $dydx$ the integral over $1$ is given by
$$int_-1^2int_x^2^x+2dydx=int_-1^2-x^2+x+2dx=[frac13x^3+frac12x^2+2x]_-1^2=frac92$$
with the region that you correctly evaluated. But for the second integral we need to split the region into two parts - for $0le yle1$ we have that $-sqrtyle xlesqrty$ and when $1le yle4$ we have $y-2le xlesqrty$. So the integral over the function $1$ is
$$int_0^1int_-sqrty^sqrtydxdy+int_1^4int_y-2^sqrtydxdy=int_0^12sqrty,dy+int_1^4sqrty-y+2,dy$$
$$=[frac43y^frac32]_0^1+[frac23y^frac32-frac12y^2+2y]_1^4=frac92$$
So the two regions are now equal.
answered Mar 30 at 19:12
Peter ForemanPeter Foreman
6,9401318
$endgroup$
For $dydx$ the integral over $1$ is given by
$$int_-1^2int_x^2^x+2dydx=int_-1^2-x^2+x+2dx=[frac13x^3+frac12x^2+2x]_-1^2=frac92$$
with the region that you correctly evaluated. But for the second integral we need to split the region into two parts - for $0le yle1$ we have that $-sqrtyle xlesqrty$ and when $1le yle4$ we have $y-2le xlesqrty$. So the integral over the function $1$ is
$$int_0^1int_-sqrty^sqrtydxdy+int_1^4int_y-2^sqrtydxdy=int_0^12sqrty,dy+int_1^4sqrty-y+2,dy$$
$$=[frac43y^frac32]_0^1+[frac23y^frac32-frac12y^2+2y]_1^4=frac92$$
So the two regions are now equal.
answered Mar 30 at 19:12
Peter ForemanPeter Foreman
6,9401318
answered Mar 30 at 19:12
Peter ForemanPeter Foreman
6,9401318
answered Mar 30 at 19:12
Peter ForemanPeter Foreman
6,9401318
answered Mar 30 at 19:12
Peter ForemanPeter Foreman
6,9401318
6,9401318
$begingroup$
Thank you this helps loads. My text book does not describe splitting them up and i did not think of it. Much appreciated!
$endgroup$
– Shaun Weinberg
Mar 31 at 10:36
add a comment |
$begingroup$
Thank you this helps loads. My text book does not describe splitting them up and i did not think of it. Much appreciated!
$endgroup$
– Shaun Weinberg
Mar 31 at 10:36
$begingroup$
Thank you this helps loads. My text book does not describe splitting them up and i did not think of it. Much appreciated!
$endgroup$
– Shaun Weinberg
Mar 31 at 10:36
$begingroup$
Thank you this helps loads. My text book does not describe splitting them up and i did not think of it. Much appreciated!
$endgroup$
– Shaun Weinberg
Mar 31 at 10:36
add a comment |
$begingroup$
For the region $H$, the lower limit of $x$ shouldn't be $y-2$.
It should be the maximum of $-sqrty$ and $y-2$. In fact, when $0 le y le 1$, the lower limit is $-sqrty$.
That is
$$H =(x,y): 0 le y le 4 , max(-sqrty, y-2) < x < sqrty $$
That is from the picture below, the left limit of the region consists of the green color and blue color part.
answered Mar 30 at 19:09
Siong Thye GohSiong Thye Goh
104k1468120
$endgroup$
$begingroup$
I originally wrote an answer for the duplicate, so I copied it here. (+1) btw
$endgroup$
– Peter Foreman
Mar 30 at 19:13
$begingroup$
I did the same thing. ;)
$endgroup$
– Siong Thye Goh
Mar 30 at 19:14
add a comment |
$begingroup$
For the region $H$, the lower limit of $x$ shouldn't be $y-2$.
It should be the maximum of $-sqrty$ and $y-2$. In fact, when $0 le y le 1$, the lower limit is $-sqrty$.
That is
$$H =(x,y): 0 le y le 4 , max(-sqrty, y-2) < x < sqrty $$
That is from the picture below, the left limit of the region consists of the green color and blue color part.
answered Mar 30 at 19:09
Siong Thye GohSiong Thye Goh
104k1468120
$endgroup$
$begingroup$
I originally wrote an answer for the duplicate, so I copied it here. (+1) btw
$endgroup$
– Peter Foreman
Mar 30 at 19:13
$begingroup$
I did the same thing. ;)
$endgroup$
– Siong Thye Goh
Mar 30 at 19:14
add a comment |
$begingroup$
For the region $H$, the lower limit of $x$ shouldn't be $y-2$.
It should be the maximum of $-sqrty$ and $y-2$. In fact, when $0 le y le 1$, the lower limit is $-sqrty$.
That is
$$H =(x,y): 0 le y le 4 , max(-sqrty, y-2) < x < sqrty $$
That is from the picture below, the left limit of the region consists of the green color and blue color part.
answered Mar 30 at 19:09
Siong Thye GohSiong Thye Goh
104k1468120
$endgroup$
For the region $H$, the lower limit of $x$ shouldn't be $y-2$.
It should be the maximum of $-sqrty$ and $y-2$. In fact, when $0 le y le 1$, the lower limit is $-sqrty$.
That is
$$H =(x,y): 0 le y le 4 , max(-sqrty, y-2) < x < sqrty $$
That is from the picture below, the left limit of the region consists of the green color and blue color part.
answered Mar 30 at 19:09
Siong Thye GohSiong Thye Goh
104k1468120
edited Mar 30 at 19:15
answered Mar 30 at 19:09
Siong Thye GohSiong Thye Goh
104k1468120
answered Mar 30 at 19:09
Siong Thye GohSiong Thye Goh
104k1468120
answered Mar 30 at 19:09
Siong Thye GohSiong Thye Goh
104k1468120
104k1468120
$begingroup$
I originally wrote an answer for the duplicate, so I copied it here. (+1) btw
$endgroup$
– Peter Foreman
Mar 30 at 19:13
$begingroup$
I did the same thing. ;)
$endgroup$
– Siong Thye Goh
Mar 30 at 19:14
add a comment |
$begingroup$
I originally wrote an answer for the duplicate, so I copied it here. (+1) btw
$endgroup$
– Peter Foreman
Mar 30 at 19:13
$begingroup$
I did the same thing. ;)
$endgroup$
– Siong Thye Goh
Mar 30 at 19:14
$begingroup$
I originally wrote an answer for the duplicate, so I copied it here. (+1) btw
$endgroup$
– Peter Foreman
Mar 30 at 19:13
$begingroup$
I originally wrote an answer for the duplicate, so I copied it here. (+1) btw
$endgroup$
– Peter Foreman
Mar 30 at 19:13
$begingroup$
I did the same thing. ;)
$endgroup$
– Siong Thye Goh
Mar 30 at 19:14
$begingroup$
I did the same thing. ;)
$endgroup$
– Siong Thye Goh
Mar 30 at 19:14
add a comment |
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