Mixing
Evaluating $iint_{R} |xy| dx dy$ over the region $R$ enclosed by $x^{2} + 4y^{2} ≥1 , x^{2} + y^{2}≤1$.
$begingroup$
I am given the following integral -
$iint_{R} |xy| dx dy$ over the region $R$ enclosed by $x^{2} + 4y^{2} ≥1 , x^{2} + y^{2}≤1.$
I know how do we solve such problems, but I have a bit confusion.
The area is enclosed by an ellipse and circle. I calculated the value of integral in first quadrant by assuming $|xy|$ simply $xy$, then I multiplied it by 4 and got my answer right. However, I want to know how does this integral, when my function is $|xy|$, is different from the integral taking function as $xy$. Why I am getting same results in both cases.
calculus
asked Jan 2 at 8:21
MathsaddictMathsaddict
3008
$endgroup$
add a comment |
$begingroup$
I am given the following integral -
$iint_{R} |xy| dx dy$ over the region $R$ enclosed by $x^{2} + 4y^{2} ≥1 , x^{2} + y^{2}≤1.$
I know how do we solve such problems, but I have a bit confusion.
The area is enclosed by an ellipse and circle. I calculated the value of integral in first quadrant by assuming $|xy|$ simply $xy$, then I multiplied it by 4 and got my answer right. However, I want to know how does this integral, when my function is $|xy|$, is different from the integral taking function as $xy$. Why I am getting same results in both cases.
calculus
asked Jan 2 at 8:21
MathsaddictMathsaddict
3008
$endgroup$
1
$begingroup$
Consider simpler example: compare $int_{-1}^1 x dx$ and $int_{-1}^1 |x| dx$.
$endgroup$
– farruhota
Jan 2 at 11:16
add a comment |
$begingroup$
I am given the following integral -
$iint_{R} |xy| dx dy$ over the region $R$ enclosed by $x^{2} + 4y^{2} ≥1 , x^{2} + y^{2}≤1.$
I know how do we solve such problems, but I have a bit confusion.
The area is enclosed by an ellipse and circle. I calculated the value of integral in first quadrant by assuming $|xy|$ simply $xy$, then I multiplied it by 4 and got my answer right. However, I want to know how does this integral, when my function is $|xy|$, is different from the integral taking function as $xy$. Why I am getting same results in both cases.
calculus
asked Jan 2 at 8:21
MathsaddictMathsaddict
3008
$endgroup$
I am given the following integral -
$iint_{R} |xy| dx dy$ over the region $R$ enclosed by $x^{2} + 4y^{2} ≥1 , x^{2} + y^{2}≤1.$
I know how do we solve such problems, but I have a bit confusion.
The area is enclosed by an ellipse and circle. I calculated the value of integral in first quadrant by assuming $|xy|$ simply $xy$, then I multiplied it by 4 and got my answer right. However, I want to know how does this integral, when my function is $|xy|$, is different from the integral taking function as $xy$. Why I am getting same results in both cases.
calculus
calculus
asked Jan 2 at 8:21
MathsaddictMathsaddict
3008
asked Jan 2 at 8:21
MathsaddictMathsaddict
3008
asked Jan 2 at 8:21
MathsaddictMathsaddict
3008
asked Jan 2 at 8:21
MathsaddictMathsaddict
3008
asked Jan 2 at 8:21
MathsaddictMathsaddict
3008
3008
1
$begingroup$
Consider simpler example: compare $int_{-1}^1 x dx$ and $int_{-1}^1 |x| dx$.
$endgroup$
– farruhota
Jan 2 at 11:16
add a comment |
1
$begingroup$
Consider simpler example: compare $int_{-1}^1 x dx$ and $int_{-1}^1 |x| dx$.
$endgroup$
– farruhota
Jan 2 at 11:16
1
1
$begingroup$
Consider simpler example: compare $int_{-1}^1 x dx$ and $int_{-1}^1 |x| dx$.
$endgroup$
– farruhota
Jan 2 at 11:16
$begingroup$
Consider simpler example: compare $int_{-1}^1 x dx$ and $int_{-1}^1 |x| dx$.
$endgroup$
– farruhota
Jan 2 at 11:16
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If there is no absolute value, by symmetry, the integral in the first quadrant is equal to the integral of the third quadrant (as $xy = (-x)(-y)$). However, the integral in the second and the fourth is going to be its negative.
Hence, they will cancel out and the answer for the problem without the absolute value sign is $0$.
edited Jan 2 at 8:35
Sauhard Sharma
890117
answered Jan 2 at 8:32
Siong Thye GohSiong Thye Goh
100k1465117
$endgroup$
add a comment |
$begingroup$
In the first quadrant it is the same, as x and y are both positive and so the absolute value doesn't do anything. This is true in the third quadrant as well as x and y are both negative and so xy is positive. But in the second and fourth quadrants $|xy| = -xy$, so it matters if the absolute value signs are there. If you integrate xy over this area than the integral is the same in the first and third quadrants, and negative that value in the second and fourth quadrants, giving an answer of 0.
answered Jan 2 at 8:33
Erik ParkinsonErik Parkinson
8749
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If there is no absolute value, by symmetry, the integral in the first quadrant is equal to the integral of the third quadrant (as $xy = (-x)(-y)$). However, the integral in the second and the fourth is going to be its negative.
Hence, they will cancel out and the answer for the problem without the absolute value sign is $0$.
edited Jan 2 at 8:35
Sauhard Sharma
890117
answered Jan 2 at 8:32
Siong Thye GohSiong Thye Goh
100k1465117
$endgroup$
add a comment |
$begingroup$
If there is no absolute value, by symmetry, the integral in the first quadrant is equal to the integral of the third quadrant (as $xy = (-x)(-y)$). However, the integral in the second and the fourth is going to be its negative.
Hence, they will cancel out and the answer for the problem without the absolute value sign is $0$.
edited Jan 2 at 8:35
Sauhard Sharma
890117
answered Jan 2 at 8:32
Siong Thye GohSiong Thye Goh
100k1465117
$endgroup$
add a comment |
$begingroup$
If there is no absolute value, by symmetry, the integral in the first quadrant is equal to the integral of the third quadrant (as $xy = (-x)(-y)$). However, the integral in the second and the fourth is going to be its negative.
Hence, they will cancel out and the answer for the problem without the absolute value sign is $0$.
edited Jan 2 at 8:35
Sauhard Sharma
890117
answered Jan 2 at 8:32
Siong Thye GohSiong Thye Goh
100k1465117
$endgroup$
If there is no absolute value, by symmetry, the integral in the first quadrant is equal to the integral of the third quadrant (as $xy = (-x)(-y)$). However, the integral in the second and the fourth is going to be its negative.
Hence, they will cancel out and the answer for the problem without the absolute value sign is $0$.
edited Jan 2 at 8:35
Sauhard Sharma
890117
answered Jan 2 at 8:32
Siong Thye GohSiong Thye Goh
100k1465117
edited Jan 2 at 8:35
Sauhard Sharma
890117
edited Jan 2 at 8:35
Sauhard Sharma
890117
edited Jan 2 at 8:35
Sauhard Sharma
890117
890117
answered Jan 2 at 8:32
Siong Thye GohSiong Thye Goh
100k1465117
answered Jan 2 at 8:32
Siong Thye GohSiong Thye Goh
100k1465117
answered Jan 2 at 8:32
Siong Thye GohSiong Thye Goh
100k1465117
100k1465117
add a comment |
add a comment |
$begingroup$
In the first quadrant it is the same, as x and y are both positive and so the absolute value doesn't do anything. This is true in the third quadrant as well as x and y are both negative and so xy is positive. But in the second and fourth quadrants $|xy| = -xy$, so it matters if the absolute value signs are there. If you integrate xy over this area than the integral is the same in the first and third quadrants, and negative that value in the second and fourth quadrants, giving an answer of 0.
answered Jan 2 at 8:33
Erik ParkinsonErik Parkinson
8749
$endgroup$
add a comment |
$begingroup$
In the first quadrant it is the same, as x and y are both positive and so the absolute value doesn't do anything. This is true in the third quadrant as well as x and y are both negative and so xy is positive. But in the second and fourth quadrants $|xy| = -xy$, so it matters if the absolute value signs are there. If you integrate xy over this area than the integral is the same in the first and third quadrants, and negative that value in the second and fourth quadrants, giving an answer of 0.
answered Jan 2 at 8:33
Erik ParkinsonErik Parkinson
8749
$endgroup$
add a comment |
$begingroup$
In the first quadrant it is the same, as x and y are both positive and so the absolute value doesn't do anything. This is true in the third quadrant as well as x and y are both negative and so xy is positive. But in the second and fourth quadrants $|xy| = -xy$, so it matters if the absolute value signs are there. If you integrate xy over this area than the integral is the same in the first and third quadrants, and negative that value in the second and fourth quadrants, giving an answer of 0.
answered Jan 2 at 8:33
Erik ParkinsonErik Parkinson
8749
$endgroup$
In the first quadrant it is the same, as x and y are both positive and so the absolute value doesn't do anything. This is true in the third quadrant as well as x and y are both negative and so xy is positive. But in the second and fourth quadrants $|xy| = -xy$, so it matters if the absolute value signs are there. If you integrate xy over this area than the integral is the same in the first and third quadrants, and negative that value in the second and fourth quadrants, giving an answer of 0.
answered Jan 2 at 8:33
Erik ParkinsonErik Parkinson
8749
answered Jan 2 at 8:33
Erik ParkinsonErik Parkinson
8749
answered Jan 2 at 8:33
Erik ParkinsonErik Parkinson
8749
answered Jan 2 at 8:33
Erik ParkinsonErik Parkinson
8749
8749
add a comment |
add a comment |
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1
$begingroup$
Consider simpler example: compare $int_{-1}^1 x dx$ and $int_{-1}^1 |x| dx$.
$endgroup$
– farruhota
Jan 2 at 11:16