Integral of multivariable function
$begingroup$
I have function $f(x,y) = x^{2} + y^{2}$ that has condition $ x^2 + y^2 le 4$.
I am suppose to create Integral definition from it.
Now i am not sure if u understood correctly how to read the boundaries.
The integral is going to be $iint left( x^2 + y^2right) dy dx$ or different order of integration ( by x then by y ).
However my question is about boundaries, we know that $x^2+y^2 le 4$, we can change it to $y = sqrt{4-x^2}$ that is the upper boundary of integration by y.
That would make $$int_a^b int_c^sqrt{4-x^2} ( x^2 + y^2) dy dx$$
However how to decide a ,b and c? I am having trouble deciding this trivial problem..
The intuition tells that minimal value of y is 0, that way value of x can be 0 to 2 which would make:
$$int_0^2 int_0^sqrt{4-x^2} ( x^2 + y^2) dy dx$$
However this is quite simple case so i am not sure how to proceed or think about it in more complex cases
Thanks for help!
calculus integration multivariable-calculus
edited Dec 12 '18 at 16:22
gt6989b
35.3k22557
asked Dec 12 '18 at 16:17
trolkuratrolkura
24519
$endgroup$
add a comment |
$begingroup$
I have function $f(x,y) = x^{2} + y^{2}$ that has condition $ x^2 + y^2 le 4$.
I am suppose to create Integral definition from it.
Now i am not sure if u understood correctly how to read the boundaries.
The integral is going to be $iint left( x^2 + y^2right) dy dx$ or different order of integration ( by x then by y ).
However my question is about boundaries, we know that $x^2+y^2 le 4$, we can change it to $y = sqrt{4-x^2}$ that is the upper boundary of integration by y.
That would make $$int_a^b int_c^sqrt{4-x^2} ( x^2 + y^2) dy dx$$
However how to decide a ,b and c? I am having trouble deciding this trivial problem..
The intuition tells that minimal value of y is 0, that way value of x can be 0 to 2 which would make:
$$int_0^2 int_0^sqrt{4-x^2} ( x^2 + y^2) dy dx$$
However this is quite simple case so i am not sure how to proceed or think about it in more complex cases
Thanks for help!
calculus integration multivariable-calculus
edited Dec 12 '18 at 16:22
gt6989b
35.3k22557
asked Dec 12 '18 at 16:17
trolkuratrolkura
24519
$endgroup$
add a comment |
$begingroup$
I have function $f(x,y) = x^{2} + y^{2}$ that has condition $ x^2 + y^2 le 4$.
I am suppose to create Integral definition from it.
Now i am not sure if u understood correctly how to read the boundaries.
The integral is going to be $iint left( x^2 + y^2right) dy dx$ or different order of integration ( by x then by y ).
However my question is about boundaries, we know that $x^2+y^2 le 4$, we can change it to $y = sqrt{4-x^2}$ that is the upper boundary of integration by y.
That would make $$int_a^b int_c^sqrt{4-x^2} ( x^2 + y^2) dy dx$$
However how to decide a ,b and c? I am having trouble deciding this trivial problem..
The intuition tells that minimal value of y is 0, that way value of x can be 0 to 2 which would make:
$$int_0^2 int_0^sqrt{4-x^2} ( x^2 + y^2) dy dx$$
However this is quite simple case so i am not sure how to proceed or think about it in more complex cases
Thanks for help!
calculus integration multivariable-calculus
edited Dec 12 '18 at 16:22
gt6989b
35.3k22557
asked Dec 12 '18 at 16:17
trolkuratrolkura
24519
$endgroup$
I have function $f(x,y) = x^{2} + y^{2}$ that has condition $ x^2 + y^2 le 4$.
I am suppose to create Integral definition from it.
Now i am not sure if u understood correctly how to read the boundaries.
The integral is going to be $iint left( x^2 + y^2right) dy dx$ or different order of integration ( by x then by y ).
However my question is about boundaries, we know that $x^2+y^2 le 4$, we can change it to $y = sqrt{4-x^2}$ that is the upper boundary of integration by y.
That would make $$int_a^b int_c^sqrt{4-x^2} ( x^2 + y^2) dy dx$$
However how to decide a ,b and c? I am having trouble deciding this trivial problem..
The intuition tells that minimal value of y is 0, that way value of x can be 0 to 2 which would make:
$$int_0^2 int_0^sqrt{4-x^2} ( x^2 + y^2) dy dx$$
However this is quite simple case so i am not sure how to proceed or think about it in more complex cases
Thanks for help!
calculus integration multivariable-calculus
calculus integration multivariable-calculus
edited Dec 12 '18 at 16:22
gt6989b
35.3k22557
asked Dec 12 '18 at 16:17
trolkuratrolkura
24519
edited Dec 12 '18 at 16:22
gt6989b
35.3k22557
asked Dec 12 '18 at 16:17
trolkuratrolkura
24519
edited Dec 12 '18 at 16:22
gt6989b
35.3k22557
edited Dec 12 '18 at 16:22
gt6989b
35.3k22557
edited Dec 12 '18 at 16:22
gt6989b
35.3k22557
35.3k22557
asked Dec 12 '18 at 16:17
trolkuratrolkura
24519
asked Dec 12 '18 at 16:17
trolkuratrolkura
24519
asked Dec 12 '18 at 16:17
trolkuratrolkura
24519
24519
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Note that it can be useful to switch to radial coordinates:
$$int_{x^2+y^2<4} (x^2+y^2) dxdy = int_{theta=0}^{2pi} int_{r=0}^{r=2} r^2 r dr dtheta$$
Edit
correction after comment below
answered Dec 12 '18 at 16:31
SebapiSebapi
515
$endgroup$
1
$begingroup$
You forgot the extra $r$ from $dx,dy = r , dr , dtheta$.
$endgroup$
– Hans Lundmark
Dec 12 '18 at 16:34
add a comment |
$begingroup$
In the region ${(x,y)inmathbb{R}^2,|,x^2+y^2leqslant4}$, the smallest possible value for $x$ is $-2$ and the largest one is $2$. And, for each $x$ in that range, the smallest valu for $y$ is $-sqrt{4-x^2}$, whereas the largest one is $sqrt{4-x^2}$.
So, an answer is$$int_{-2}^2int_{-sqrt{4-x^2}}^{sqrt{4-x^2}}x^2+y^2,mathrm dy,mathrm dx.$$
answered Dec 12 '18 at 16:21
José Carlos SantosJosé Carlos Santos
172k22132239
$endgroup$
add a comment |
$begingroup$
Here is how I think about it.
First I sketch my region. In this case it is pretty simple. It is a disk of radius 2.
Decide whether it looks like it will be easier to integrate by $y$ first or by $x$ first. Generally, you will pick the one that does not require you to break up the integration.
In this case, it doesn't make a difference. So, lets integrate by $y$ first.
We need the limits for $y.$ Draw a vertical line through your region, the endpoints of this line will indicate the limits for $y.$
If we were integrating by $x$ first it would be a horizontal line.
What will be the formula for those points?
$-sqrt {4-x^2}, sqrt {4-x^2}$
Seeing it like this gets you out of the trap of guessing that the lower limit might be $0,$ or something similar.
How far can $x$ rage before you bust the limits of your region?
$int_{-2}^2 int_{-sqrt{4-x^2}}^{sqrt{4-x^2}} f(x,y) dy dx$
When you have more experience, you can trust your intuition. Until then, don't.
Of course the easy way to do this is to convert to polar, but that is probably beyond your level right now.
answered Dec 12 '18 at 18:46
Doug MDoug M
45.3k31954
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Note that it can be useful to switch to radial coordinates:
$$int_{x^2+y^2<4} (x^2+y^2) dxdy = int_{theta=0}^{2pi} int_{r=0}^{r=2} r^2 r dr dtheta$$
Edit
correction after comment below
answered Dec 12 '18 at 16:31
SebapiSebapi
515
$endgroup$
1
$begingroup$
You forgot the extra $r$ from $dx,dy = r , dr , dtheta$.
$endgroup$
– Hans Lundmark
Dec 12 '18 at 16:34
add a comment |
$begingroup$
Note that it can be useful to switch to radial coordinates:
$$int_{x^2+y^2<4} (x^2+y^2) dxdy = int_{theta=0}^{2pi} int_{r=0}^{r=2} r^2 r dr dtheta$$
Edit
correction after comment below
answered Dec 12 '18 at 16:31
SebapiSebapi
515
$endgroup$
1
$begingroup$
You forgot the extra $r$ from $dx,dy = r , dr , dtheta$.
$endgroup$
– Hans Lundmark
Dec 12 '18 at 16:34
add a comment |
$begingroup$
Note that it can be useful to switch to radial coordinates:
$$int_{x^2+y^2<4} (x^2+y^2) dxdy = int_{theta=0}^{2pi} int_{r=0}^{r=2} r^2 r dr dtheta$$
Edit
correction after comment below
answered Dec 12 '18 at 16:31
SebapiSebapi
515
$endgroup$
Note that it can be useful to switch to radial coordinates:
$$int_{x^2+y^2<4} (x^2+y^2) dxdy = int_{theta=0}^{2pi} int_{r=0}^{r=2} r^2 r dr dtheta$$
Edit
correction after comment below
answered Dec 12 '18 at 16:31
SebapiSebapi
515
edited Dec 12 '18 at 16:40
answered Dec 12 '18 at 16:31
SebapiSebapi
515
answered Dec 12 '18 at 16:31
SebapiSebapi
515
answered Dec 12 '18 at 16:31
SebapiSebapi
515
515
1
$begingroup$
You forgot the extra $r$ from $dx,dy = r , dr , dtheta$.
$endgroup$
– Hans Lundmark
Dec 12 '18 at 16:34
add a comment |
1
$begingroup$
You forgot the extra $r$ from $dx,dy = r , dr , dtheta$.
$endgroup$
– Hans Lundmark
Dec 12 '18 at 16:34
1
1
$begingroup$
You forgot the extra $r$ from $dx,dy = r , dr , dtheta$.
$endgroup$
– Hans Lundmark
Dec 12 '18 at 16:34
$begingroup$
You forgot the extra $r$ from $dx,dy = r , dr , dtheta$.
$endgroup$
– Hans Lundmark
Dec 12 '18 at 16:34
add a comment |
$begingroup$
In the region ${(x,y)inmathbb{R}^2,|,x^2+y^2leqslant4}$, the smallest possible value for $x$ is $-2$ and the largest one is $2$. And, for each $x$ in that range, the smallest valu for $y$ is $-sqrt{4-x^2}$, whereas the largest one is $sqrt{4-x^2}$.
So, an answer is$$int_{-2}^2int_{-sqrt{4-x^2}}^{sqrt{4-x^2}}x^2+y^2,mathrm dy,mathrm dx.$$
answered Dec 12 '18 at 16:21
José Carlos SantosJosé Carlos Santos
172k22132239
$endgroup$
add a comment |
$begingroup$
In the region ${(x,y)inmathbb{R}^2,|,x^2+y^2leqslant4}$, the smallest possible value for $x$ is $-2$ and the largest one is $2$. And, for each $x$ in that range, the smallest valu for $y$ is $-sqrt{4-x^2}$, whereas the largest one is $sqrt{4-x^2}$.
So, an answer is$$int_{-2}^2int_{-sqrt{4-x^2}}^{sqrt{4-x^2}}x^2+y^2,mathrm dy,mathrm dx.$$
answered Dec 12 '18 at 16:21
José Carlos SantosJosé Carlos Santos
172k22132239
$endgroup$
add a comment |
$begingroup$
In the region ${(x,y)inmathbb{R}^2,|,x^2+y^2leqslant4}$, the smallest possible value for $x$ is $-2$ and the largest one is $2$. And, for each $x$ in that range, the smallest valu for $y$ is $-sqrt{4-x^2}$, whereas the largest one is $sqrt{4-x^2}$.
So, an answer is$$int_{-2}^2int_{-sqrt{4-x^2}}^{sqrt{4-x^2}}x^2+y^2,mathrm dy,mathrm dx.$$
answered Dec 12 '18 at 16:21
José Carlos SantosJosé Carlos Santos
172k22132239
$endgroup$
In the region ${(x,y)inmathbb{R}^2,|,x^2+y^2leqslant4}$, the smallest possible value for $x$ is $-2$ and the largest one is $2$. And, for each $x$ in that range, the smallest valu for $y$ is $-sqrt{4-x^2}$, whereas the largest one is $sqrt{4-x^2}$.
So, an answer is$$int_{-2}^2int_{-sqrt{4-x^2}}^{sqrt{4-x^2}}x^2+y^2,mathrm dy,mathrm dx.$$
answered Dec 12 '18 at 16:21
José Carlos SantosJosé Carlos Santos
172k22132239
edited Dec 12 '18 at 18:13
answered Dec 12 '18 at 16:21
José Carlos SantosJosé Carlos Santos
172k22132239
answered Dec 12 '18 at 16:21
José Carlos SantosJosé Carlos Santos
172k22132239
answered Dec 12 '18 at 16:21
José Carlos SantosJosé Carlos Santos
172k22132239
172k22132239
add a comment |
add a comment |
$begingroup$
Here is how I think about it.
First I sketch my region. In this case it is pretty simple. It is a disk of radius 2.
Decide whether it looks like it will be easier to integrate by $y$ first or by $x$ first. Generally, you will pick the one that does not require you to break up the integration.
In this case, it doesn't make a difference. So, lets integrate by $y$ first.
We need the limits for $y.$ Draw a vertical line through your region, the endpoints of this line will indicate the limits for $y.$
If we were integrating by $x$ first it would be a horizontal line.
What will be the formula for those points?
$-sqrt {4-x^2}, sqrt {4-x^2}$
Seeing it like this gets you out of the trap of guessing that the lower limit might be $0,$ or something similar.
How far can $x$ rage before you bust the limits of your region?
$int_{-2}^2 int_{-sqrt{4-x^2}}^{sqrt{4-x^2}} f(x,y) dy dx$
When you have more experience, you can trust your intuition. Until then, don't.
Of course the easy way to do this is to convert to polar, but that is probably beyond your level right now.
answered Dec 12 '18 at 18:46
Doug MDoug M
45.3k31954
$endgroup$
add a comment |
$begingroup$
Here is how I think about it.
First I sketch my region. In this case it is pretty simple. It is a disk of radius 2.
Decide whether it looks like it will be easier to integrate by $y$ first or by $x$ first. Generally, you will pick the one that does not require you to break up the integration.
In this case, it doesn't make a difference. So, lets integrate by $y$ first.
We need the limits for $y.$ Draw a vertical line through your region, the endpoints of this line will indicate the limits for $y.$
If we were integrating by $x$ first it would be a horizontal line.
What will be the formula for those points?
$-sqrt {4-x^2}, sqrt {4-x^2}$
Seeing it like this gets you out of the trap of guessing that the lower limit might be $0,$ or something similar.
How far can $x$ rage before you bust the limits of your region?
$int_{-2}^2 int_{-sqrt{4-x^2}}^{sqrt{4-x^2}} f(x,y) dy dx$
When you have more experience, you can trust your intuition. Until then, don't.
Of course the easy way to do this is to convert to polar, but that is probably beyond your level right now.
answered Dec 12 '18 at 18:46
Doug MDoug M
45.3k31954
$endgroup$
add a comment |
$begingroup$
Here is how I think about it.
First I sketch my region. In this case it is pretty simple. It is a disk of radius 2.
Decide whether it looks like it will be easier to integrate by $y$ first or by $x$ first. Generally, you will pick the one that does not require you to break up the integration.
In this case, it doesn't make a difference. So, lets integrate by $y$ first.
We need the limits for $y.$ Draw a vertical line through your region, the endpoints of this line will indicate the limits for $y.$
If we were integrating by $x$ first it would be a horizontal line.
What will be the formula for those points?
$-sqrt {4-x^2}, sqrt {4-x^2}$
Seeing it like this gets you out of the trap of guessing that the lower limit might be $0,$ or something similar.
How far can $x$ rage before you bust the limits of your region?
$int_{-2}^2 int_{-sqrt{4-x^2}}^{sqrt{4-x^2}} f(x,y) dy dx$
When you have more experience, you can trust your intuition. Until then, don't.
Of course the easy way to do this is to convert to polar, but that is probably beyond your level right now.
answered Dec 12 '18 at 18:46
Doug MDoug M
45.3k31954
$endgroup$
Here is how I think about it.
First I sketch my region. In this case it is pretty simple. It is a disk of radius 2.
Decide whether it looks like it will be easier to integrate by $y$ first or by $x$ first. Generally, you will pick the one that does not require you to break up the integration.
In this case, it doesn't make a difference. So, lets integrate by $y$ first.
We need the limits for $y.$ Draw a vertical line through your region, the endpoints of this line will indicate the limits for $y.$
If we were integrating by $x$ first it would be a horizontal line.
What will be the formula for those points?
$-sqrt {4-x^2}, sqrt {4-x^2}$
Seeing it like this gets you out of the trap of guessing that the lower limit might be $0,$ or something similar.
How far can $x$ rage before you bust the limits of your region?
$int_{-2}^2 int_{-sqrt{4-x^2}}^{sqrt{4-x^2}} f(x,y) dy dx$
When you have more experience, you can trust your intuition. Until then, don't.
Of course the easy way to do this is to convert to polar, but that is probably beyond your level right now.
answered Dec 12 '18 at 18:46
Doug MDoug M
45.3k31954
answered Dec 12 '18 at 18:46
Doug MDoug M
45.3k31954
answered Dec 12 '18 at 18:46
Doug MDoug M
45.3k31954
answered Dec 12 '18 at 18:46
Doug MDoug M
45.3k31954
45.3k31954
add a comment |
add a comment |
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