Can we calculate double integrals without drawing the sketch?
0
$begingroup$
I found it very hard to draw the sketch so my question is can we solve the question numerically without drawing the sketch ?
For example I have a question
$$iint_D arctan{frac{y}{x}},dxdy$$
where $D$ is domain bounded by $x^2+y^2=1$, $x^2+y^2=4$, $y=x$, $y = sqrt{3}x$.
multivariable-calculus
edited Jan 22 at 8:54
Jyrki Lahtonen
109k13170376
asked Jan 22 at 8:19
Hamza.SHamza.S
1236
$endgroup$
|
show 4 more comments
0
$begingroup$
I found it very hard to draw the sketch so my question is can we solve the question numerically without drawing the sketch ?
For example I have a question
$$iint_D arctan{frac{y}{x}},dxdy$$
where $D$ is domain bounded by $x^2+y^2=1$, $x^2+y^2=4$, $y=x$, $y = sqrt{3}x$.
multivariable-calculus
edited Jan 22 at 8:54
Jyrki Lahtonen
109k13170376
asked Jan 22 at 8:19
Hamza.SHamza.S
1236
$endgroup$
$begingroup$
why you downvote it @Matti P. ??
$endgroup$
– Hamza.S
Jan 22 at 8:21
$begingroup$
I didn't, it was someone else
$endgroup$
– Matti P.
Jan 22 at 8:22
1
$begingroup$
I am not the downvoter, but to prevent more downvotes you should show an attempt.
$endgroup$
– Robert Z
Jan 22 at 8:24
4
$begingroup$
Solving a double integral when the limits are explicitly given is easy. Solving it on a domain requires figuring out the limits that is very difficult to do without a physical or mental sketch of the domain. In your case, however, you can simplify the integral and the domain by using polar coordinates
$endgroup$
– Shubham Johri
Jan 22 at 8:25
4
$begingroup$
But, yes, you definitely should use the polar coordinates here. And, yes, I believe in drawing pictures in cases like these. The reason is that in applications, particularly those originating in physics, the boundaries are defined visually rather than using equations. If you cannot figure out the boundaries from a given image, you absolutely must practice that process.
$endgroup$
– Jyrki Lahtonen
Jan 22 at 8:34
|
show 4 more comments
0
0
0
$begingroup$
I found it very hard to draw the sketch so my question is can we solve the question numerically without drawing the sketch ?
For example I have a question
$$iint_D arctan{frac{y}{x}},dxdy$$
where $D$ is domain bounded by $x^2+y^2=1$, $x^2+y^2=4$, $y=x$, $y = sqrt{3}x$.
multivariable-calculus
edited Jan 22 at 8:54
Jyrki Lahtonen
109k13170376
asked Jan 22 at 8:19
Hamza.SHamza.S
1236
$endgroup$
I found it very hard to draw the sketch so my question is can we solve the question numerically without drawing the sketch ?
For example I have a question
$$iint_D arctan{frac{y}{x}},dxdy$$
where $D$ is domain bounded by $x^2+y^2=1$, $x^2+y^2=4$, $y=x$, $y = sqrt{3}x$.
multivariable-calculus
multivariable-calculus
edited Jan 22 at 8:54
Jyrki Lahtonen
109k13170376
asked Jan 22 at 8:19
Hamza.SHamza.S
1236
edited Jan 22 at 8:54
Jyrki Lahtonen
109k13170376
asked Jan 22 at 8:19
Hamza.SHamza.S
1236
edited Jan 22 at 8:54
Jyrki Lahtonen
109k13170376
edited Jan 22 at 8:54
Jyrki Lahtonen
109k13170376
edited Jan 22 at 8:54
Jyrki Lahtonen
109k13170376
109k13170376
asked Jan 22 at 8:19
Hamza.SHamza.S
1236
asked Jan 22 at 8:19
Hamza.SHamza.S
1236
asked Jan 22 at 8:19
Hamza.SHamza.S
1236
1236
$begingroup$
why you downvote it @Matti P. ??
$endgroup$
– Hamza.S
Jan 22 at 8:21
$begingroup$
I didn't, it was someone else
$endgroup$
– Matti P.
Jan 22 at 8:22
1
$begingroup$
I am not the downvoter, but to prevent more downvotes you should show an attempt.
$endgroup$
– Robert Z
Jan 22 at 8:24
4
$begingroup$
Solving a double integral when the limits are explicitly given is easy. Solving it on a domain requires figuring out the limits that is very difficult to do without a physical or mental sketch of the domain. In your case, however, you can simplify the integral and the domain by using polar coordinates
$endgroup$
– Shubham Johri
Jan 22 at 8:25
4
$begingroup$
But, yes, you definitely should use the polar coordinates here. And, yes, I believe in drawing pictures in cases like these. The reason is that in applications, particularly those originating in physics, the boundaries are defined visually rather than using equations. If you cannot figure out the boundaries from a given image, you absolutely must practice that process.
$endgroup$
– Jyrki Lahtonen
Jan 22 at 8:34
|
show 4 more comments
$begingroup$
why you downvote it @Matti P. ??
$endgroup$
– Hamza.S
Jan 22 at 8:21
$begingroup$
I didn't, it was someone else
$endgroup$
– Matti P.
Jan 22 at 8:22
1
$begingroup$
I am not the downvoter, but to prevent more downvotes you should show an attempt.
$endgroup$
– Robert Z
Jan 22 at 8:24
4
$begingroup$
Solving a double integral when the limits are explicitly given is easy. Solving it on a domain requires figuring out the limits that is very difficult to do without a physical or mental sketch of the domain. In your case, however, you can simplify the integral and the domain by using polar coordinates
$endgroup$
– Shubham Johri
Jan 22 at 8:25
4
$begingroup$
But, yes, you definitely should use the polar coordinates here. And, yes, I believe in drawing pictures in cases like these. The reason is that in applications, particularly those originating in physics, the boundaries are defined visually rather than using equations. If you cannot figure out the boundaries from a given image, you absolutely must practice that process.
$endgroup$
– Jyrki Lahtonen
Jan 22 at 8:34
$begingroup$
why you downvote it @Matti P. ??
$endgroup$
– Hamza.S
Jan 22 at 8:21
$begingroup$
why you downvote it @Matti P. ??
$endgroup$
– Hamza.S
Jan 22 at 8:21
$begingroup$
I didn't, it was someone else
$endgroup$
– Matti P.
Jan 22 at 8:22
$begingroup$
I didn't, it was someone else
$endgroup$
– Matti P.
Jan 22 at 8:22
1
1
$begingroup$
I am not the downvoter, but to prevent more downvotes you should show an attempt.
$endgroup$
– Robert Z
Jan 22 at 8:24
$begingroup$
I am not the downvoter, but to prevent more downvotes you should show an attempt.
$endgroup$
– Robert Z
Jan 22 at 8:24
4
4
$begingroup$
Solving a double integral when the limits are explicitly given is easy. Solving it on a domain requires figuring out the limits that is very difficult to do without a physical or mental sketch of the domain. In your case, however, you can simplify the integral and the domain by using polar coordinates
$endgroup$
– Shubham Johri
Jan 22 at 8:25
$begingroup$
Solving a double integral when the limits are explicitly given is easy. Solving it on a domain requires figuring out the limits that is very difficult to do without a physical or mental sketch of the domain. In your case, however, you can simplify the integral and the domain by using polar coordinates
$endgroup$
– Shubham Johri
Jan 22 at 8:25
4
4
$begingroup$
But, yes, you definitely should use the polar coordinates here. And, yes, I believe in drawing pictures in cases like these. The reason is that in applications, particularly those originating in physics, the boundaries are defined visually rather than using equations. If you cannot figure out the boundaries from a given image, you absolutely must practice that process.
$endgroup$
– Jyrki Lahtonen
Jan 22 at 8:34
$begingroup$
But, yes, you definitely should use the polar coordinates here. And, yes, I believe in drawing pictures in cases like these. The reason is that in applications, particularly those originating in physics, the boundaries are defined visually rather than using equations. If you cannot figure out the boundaries from a given image, you absolutely must practice that process.
$endgroup$
– Jyrki Lahtonen
Jan 22 at 8:34
|
show 4 more comments
1 Answer
1
active
oldest
votes
2
$begingroup$
You will eventually have to make a physical or mental sketch of the domain to figure out the limits of integration. You can try to simplify the given equations through some kind of substitution. Polar coordinates work well here.
Keep $x^2+y^2=r^2,theta=arctan(y/x)$.
The domain $D$ is flanked by $r=1,r=2,theta=pi/4,theta=pi/3$. This is pretty simple to sketch.
The integral in the first quadrant becomes$$int_{pi/4}^{pi/3}int_1^2theta rdr~dtheta$$
answered Jan 22 at 8:36
Shubham JohriShubham Johri
5,189718
$endgroup$
$begingroup$
thanks . but sorry for the stupid question i am a beginner i found the sketching very hard to draw .
$endgroup$
– Hamza.S
Jan 22 at 8:38
1
$begingroup$
@Hamas.S If you are not familiar with graphing polar curves, look here.
$endgroup$
– Shubham Johri
Jan 22 at 9:00
$begingroup$
Ok Thanks @ShubhamJohri
$endgroup$
– Hamza.S
Jan 22 at 9:02
2
$begingroup$
@Hamza.S Note that I used polar coordinates for ease of integration. The integrand $arctan(y/x)$ is fairly difficult to integrate in Cartesian coordinates. In polar coordinates, it reduces to a mere $theta$ in the first quadrant. For the purpose of drawing the sketch, it is equally easy to observe that in the Cartesian system, the first two equations represent circles centred at $(0,0)$ with radii $1,2$ respectively while the the latter represent straight lines passing through $(0,0)$ and inclined at $45deg,60deg$ in the anticlockwise direction with the positive $x$ axis respectively
$endgroup$
– Shubham Johri
Jan 22 at 9:05
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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oldest
votes
2
$begingroup$
You will eventually have to make a physical or mental sketch of the domain to figure out the limits of integration. You can try to simplify the given equations through some kind of substitution. Polar coordinates work well here.
Keep $x^2+y^2=r^2,theta=arctan(y/x)$.
The domain $D$ is flanked by $r=1,r=2,theta=pi/4,theta=pi/3$. This is pretty simple to sketch.
The integral in the first quadrant becomes$$int_{pi/4}^{pi/3}int_1^2theta rdr~dtheta$$
answered Jan 22 at 8:36
Shubham JohriShubham Johri
5,189718
$endgroup$
$begingroup$
thanks . but sorry for the stupid question i am a beginner i found the sketching very hard to draw .
$endgroup$
– Hamza.S
Jan 22 at 8:38
1
$begingroup$
@Hamas.S If you are not familiar with graphing polar curves, look here.
$endgroup$
– Shubham Johri
Jan 22 at 9:00
$begingroup$
Ok Thanks @ShubhamJohri
$endgroup$
– Hamza.S
Jan 22 at 9:02
2
$begingroup$
@Hamza.S Note that I used polar coordinates for ease of integration. The integrand $arctan(y/x)$ is fairly difficult to integrate in Cartesian coordinates. In polar coordinates, it reduces to a mere $theta$ in the first quadrant. For the purpose of drawing the sketch, it is equally easy to observe that in the Cartesian system, the first two equations represent circles centred at $(0,0)$ with radii $1,2$ respectively while the the latter represent straight lines passing through $(0,0)$ and inclined at $45deg,60deg$ in the anticlockwise direction with the positive $x$ axis respectively
$endgroup$
– Shubham Johri
Jan 22 at 9:05
add a comment |
2
$begingroup$
You will eventually have to make a physical or mental sketch of the domain to figure out the limits of integration. You can try to simplify the given equations through some kind of substitution. Polar coordinates work well here.
Keep $x^2+y^2=r^2,theta=arctan(y/x)$.
The domain $D$ is flanked by $r=1,r=2,theta=pi/4,theta=pi/3$. This is pretty simple to sketch.
The integral in the first quadrant becomes$$int_{pi/4}^{pi/3}int_1^2theta rdr~dtheta$$
answered Jan 22 at 8:36
Shubham JohriShubham Johri
5,189718
$endgroup$
$begingroup$
thanks . but sorry for the stupid question i am a beginner i found the sketching very hard to draw .
$endgroup$
– Hamza.S
Jan 22 at 8:38
1
$begingroup$
@Hamas.S If you are not familiar with graphing polar curves, look here.
$endgroup$
– Shubham Johri
Jan 22 at 9:00
$begingroup$
Ok Thanks @ShubhamJohri
$endgroup$
– Hamza.S
Jan 22 at 9:02
2
$begingroup$
@Hamza.S Note that I used polar coordinates for ease of integration. The integrand $arctan(y/x)$ is fairly difficult to integrate in Cartesian coordinates. In polar coordinates, it reduces to a mere $theta$ in the first quadrant. For the purpose of drawing the sketch, it is equally easy to observe that in the Cartesian system, the first two equations represent circles centred at $(0,0)$ with radii $1,2$ respectively while the the latter represent straight lines passing through $(0,0)$ and inclined at $45deg,60deg$ in the anticlockwise direction with the positive $x$ axis respectively
$endgroup$
– Shubham Johri
Jan 22 at 9:05
add a comment |
2
2
2
$begingroup$
You will eventually have to make a physical or mental sketch of the domain to figure out the limits of integration. You can try to simplify the given equations through some kind of substitution. Polar coordinates work well here.
Keep $x^2+y^2=r^2,theta=arctan(y/x)$.
The domain $D$ is flanked by $r=1,r=2,theta=pi/4,theta=pi/3$. This is pretty simple to sketch.
The integral in the first quadrant becomes$$int_{pi/4}^{pi/3}int_1^2theta rdr~dtheta$$
answered Jan 22 at 8:36
Shubham JohriShubham Johri
5,189718
$endgroup$
You will eventually have to make a physical or mental sketch of the domain to figure out the limits of integration. You can try to simplify the given equations through some kind of substitution. Polar coordinates work well here.
Keep $x^2+y^2=r^2,theta=arctan(y/x)$.
The domain $D$ is flanked by $r=1,r=2,theta=pi/4,theta=pi/3$. This is pretty simple to sketch.
The integral in the first quadrant becomes$$int_{pi/4}^{pi/3}int_1^2theta rdr~dtheta$$
answered Jan 22 at 8:36
Shubham JohriShubham Johri
5,189718
edited Jan 22 at 9:13
answered Jan 22 at 8:36
Shubham JohriShubham Johri
5,189718
answered Jan 22 at 8:36
Shubham JohriShubham Johri
5,189718
answered Jan 22 at 8:36
Shubham JohriShubham Johri
5,189718
5,189718
$begingroup$
thanks . but sorry for the stupid question i am a beginner i found the sketching very hard to draw .
$endgroup$
– Hamza.S
Jan 22 at 8:38
1
$begingroup$
@Hamas.S If you are not familiar with graphing polar curves, look here.
$endgroup$
– Shubham Johri
Jan 22 at 9:00
$begingroup$
Ok Thanks @ShubhamJohri
$endgroup$
– Hamza.S
Jan 22 at 9:02
2
$begingroup$
@Hamza.S Note that I used polar coordinates for ease of integration. The integrand $arctan(y/x)$ is fairly difficult to integrate in Cartesian coordinates. In polar coordinates, it reduces to a mere $theta$ in the first quadrant. For the purpose of drawing the sketch, it is equally easy to observe that in the Cartesian system, the first two equations represent circles centred at $(0,0)$ with radii $1,2$ respectively while the the latter represent straight lines passing through $(0,0)$ and inclined at $45deg,60deg$ in the anticlockwise direction with the positive $x$ axis respectively
$endgroup$
– Shubham Johri
Jan 22 at 9:05
add a comment |
$begingroup$
thanks . but sorry for the stupid question i am a beginner i found the sketching very hard to draw .
$endgroup$
– Hamza.S
Jan 22 at 8:38
1
$begingroup$
@Hamas.S If you are not familiar with graphing polar curves, look here.
$endgroup$
– Shubham Johri
Jan 22 at 9:00
$begingroup$
Ok Thanks @ShubhamJohri
$endgroup$
– Hamza.S
Jan 22 at 9:02
2
$begingroup$
@Hamza.S Note that I used polar coordinates for ease of integration. The integrand $arctan(y/x)$ is fairly difficult to integrate in Cartesian coordinates. In polar coordinates, it reduces to a mere $theta$ in the first quadrant. For the purpose of drawing the sketch, it is equally easy to observe that in the Cartesian system, the first two equations represent circles centred at $(0,0)$ with radii $1,2$ respectively while the the latter represent straight lines passing through $(0,0)$ and inclined at $45deg,60deg$ in the anticlockwise direction with the positive $x$ axis respectively
$endgroup$
– Shubham Johri
Jan 22 at 9:05
$begingroup$
thanks . but sorry for the stupid question i am a beginner i found the sketching very hard to draw .
$endgroup$
– Hamza.S
Jan 22 at 8:38
$begingroup$
thanks . but sorry for the stupid question i am a beginner i found the sketching very hard to draw .
$endgroup$
– Hamza.S
Jan 22 at 8:38
1
1
$begingroup$
@Hamas.S If you are not familiar with graphing polar curves, look here.
$endgroup$
– Shubham Johri
Jan 22 at 9:00
$begingroup$
@Hamas.S If you are not familiar with graphing polar curves, look here.
$endgroup$
– Shubham Johri
Jan 22 at 9:00
$begingroup$
Ok Thanks @ShubhamJohri
$endgroup$
– Hamza.S
Jan 22 at 9:02
$begingroup$
Ok Thanks @ShubhamJohri
$endgroup$
– Hamza.S
Jan 22 at 9:02
2
2
$begingroup$
@Hamza.S Note that I used polar coordinates for ease of integration. The integrand $arctan(y/x)$ is fairly difficult to integrate in Cartesian coordinates. In polar coordinates, it reduces to a mere $theta$ in the first quadrant. For the purpose of drawing the sketch, it is equally easy to observe that in the Cartesian system, the first two equations represent circles centred at $(0,0)$ with radii $1,2$ respectively while the the latter represent straight lines passing through $(0,0)$ and inclined at $45deg,60deg$ in the anticlockwise direction with the positive $x$ axis respectively
$endgroup$
– Shubham Johri
Jan 22 at 9:05
$begingroup$
@Hamza.S Note that I used polar coordinates for ease of integration. The integrand $arctan(y/x)$ is fairly difficult to integrate in Cartesian coordinates. In polar coordinates, it reduces to a mere $theta$ in the first quadrant. For the purpose of drawing the sketch, it is equally easy to observe that in the Cartesian system, the first two equations represent circles centred at $(0,0)$ with radii $1,2$ respectively while the the latter represent straight lines passing through $(0,0)$ and inclined at $45deg,60deg$ in the anticlockwise direction with the positive $x$ axis respectively
$endgroup$
– Shubham Johri
Jan 22 at 9:05
add a comment |
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$begingroup$
why you downvote it @Matti P. ??
$endgroup$
– Hamza.S
Jan 22 at 8:21
$begingroup$
I didn't, it was someone else
$endgroup$
– Matti P.
Jan 22 at 8:22
1
$begingroup$
I am not the downvoter, but to prevent more downvotes you should show an attempt.
$endgroup$
– Robert Z
Jan 22 at 8:24
4
$begingroup$
Solving a double integral when the limits are explicitly given is easy. Solving it on a domain requires figuring out the limits that is very difficult to do without a physical or mental sketch of the domain. In your case, however, you can simplify the integral and the domain by using polar coordinates
$endgroup$
– Shubham Johri
Jan 22 at 8:25
4
$begingroup$
But, yes, you definitely should use the polar coordinates here. And, yes, I believe in drawing pictures in cases like these. The reason is that in applications, particularly those originating in physics, the boundaries are defined visually rather than using equations. If you cannot figure out the boundaries from a given image, you absolutely must practice that process.
$endgroup$
– Jyrki Lahtonen
Jan 22 at 8:34