Mixing
Finding region of double integral [closed]
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How can i calculate the double integral of $x^2+y^2$ which is bounded between the lines $x=1 , x=2$ and $y=x , y=x^{1/2}$ ? Should i look at the region between the graphs of the above functions ?
definite-integrals
edited Jan 16 at 2:33
Key Flex
8,28261233
asked Jan 16 at 2:14
user599310user599310
1071
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closed as off-topic by Cesareo, Lord_Farin, user91500, Adrian Keister, Arnaud D. Jan 16 at 23:22
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Cesareo, Lord_Farin, user91500, Adrian Keister, Arnaud D.
If this question can be reworded to fit the rules in the help center, please edit the question.
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$begingroup$
How can i calculate the double integral of $x^2+y^2$ which is bounded between the lines $x=1 , x=2$ and $y=x , y=x^{1/2}$ ? Should i look at the region between the graphs of the above functions ?
definite-integrals
edited Jan 16 at 2:33
Key Flex
8,28261233
asked Jan 16 at 2:14
user599310user599310
1071
$endgroup$
closed as off-topic by Cesareo, Lord_Farin, user91500, Adrian Keister, Arnaud D. Jan 16 at 23:22
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Cesareo, Lord_Farin, user91500, Adrian Keister, Arnaud D.
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
How can i calculate the double integral of $x^2+y^2$ which is bounded between the lines $x=1 , x=2$ and $y=x , y=x^{1/2}$ ? Should i look at the region between the graphs of the above functions ?
definite-integrals
edited Jan 16 at 2:33
Key Flex
8,28261233
asked Jan 16 at 2:14
user599310user599310
1071
$endgroup$
How can i calculate the double integral of $x^2+y^2$ which is bounded between the lines $x=1 , x=2$ and $y=x , y=x^{1/2}$ ? Should i look at the region between the graphs of the above functions ?
definite-integrals
definite-integrals
edited Jan 16 at 2:33
Key Flex
8,28261233
asked Jan 16 at 2:14
user599310user599310
1071
edited Jan 16 at 2:33
Key Flex
8,28261233
asked Jan 16 at 2:14
user599310user599310
1071
edited Jan 16 at 2:33
Key Flex
8,28261233
edited Jan 16 at 2:33
Key Flex
8,28261233
edited Jan 16 at 2:33
Key Flex
8,28261233
8,28261233
asked Jan 16 at 2:14
user599310user599310
1071
asked Jan 16 at 2:14
user599310user599310
1071
asked Jan 16 at 2:14
user599310user599310
1071
1071
closed as off-topic by Cesareo, Lord_Farin, user91500, Adrian Keister, Arnaud D. Jan 16 at 23:22
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Cesareo, Lord_Farin, user91500, Adrian Keister, Arnaud D.
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Cesareo, Lord_Farin, user91500, Adrian Keister, Arnaud D. Jan 16 at 23:22
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Cesareo, Lord_Farin, user91500, Adrian Keister, Arnaud D.
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
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We have a function in two variables $f(x,y)=x^2+y^2$ over the region bounded by the curves $x=1$, $x=2$, $y=x$, and $y=x^{1/2}$. Here is a Desmos plot of these curves:
The only region bounded by all four curves seems to be the curved triangle containing $(1.5,1.5)$. To set up the integral find lower and upper bounds on both variables. It is $1leq xleq2$ and $x^{1/2}<y<x$. Then we have the double integral:
$$I=int_{x=1}^2int_{y=x^{1/2}}^x x^2+y^2text{ }dytext{ }dx$$
First take the antiderivative of $f(x,y)$ with respect to $y$:
$$=int_{x=1}^2Big[x^2y+frac{1}{3}y^3Big]_{y=x^{1/2}}^xdx$$
Evaluate from $y=x^{1/2}$ to $x$:
$$=int_{x=1}^2Big[Big(x^3+frac{1}{3}x^3Big)-Big(x^{5/2}+frac{1}{3}x^{3/2}Big)Big]dx$$
$$=int_{x=1}^2frac{4}{3}x^3-x^{5/2}-frac{1}{3}x^{3/2}text{ }dx$$
answered Jan 16 at 2:36
M. NestorM. Nestor
778113
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$begingroup$
Isnt the upper limit x ?
$endgroup$
– user599310
Jan 16 at 3:37
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Yes, the lower bound is $y_min=x^{1/2}$ and the upper bound is $y_max=x$.
$endgroup$
– M. Nestor
Jan 16 at 3:44
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i am trying to do the calculations but are too hard
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– user599310
Jan 18 at 22:32
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I edited my answer, perhaps you can complete the calculation.
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– M. Nestor
Jan 18 at 23:56
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We have a function in two variables $f(x,y)=x^2+y^2$ over the region bounded by the curves $x=1$, $x=2$, $y=x$, and $y=x^{1/2}$. Here is a Desmos plot of these curves:
The only region bounded by all four curves seems to be the curved triangle containing $(1.5,1.5)$. To set up the integral find lower and upper bounds on both variables. It is $1leq xleq2$ and $x^{1/2}<y<x$. Then we have the double integral:
$$I=int_{x=1}^2int_{y=x^{1/2}}^x x^2+y^2text{ }dytext{ }dx$$
First take the antiderivative of $f(x,y)$ with respect to $y$:
$$=int_{x=1}^2Big[x^2y+frac{1}{3}y^3Big]_{y=x^{1/2}}^xdx$$
Evaluate from $y=x^{1/2}$ to $x$:
$$=int_{x=1}^2Big[Big(x^3+frac{1}{3}x^3Big)-Big(x^{5/2}+frac{1}{3}x^{3/2}Big)Big]dx$$
$$=int_{x=1}^2frac{4}{3}x^3-x^{5/2}-frac{1}{3}x^{3/2}text{ }dx$$
answered Jan 16 at 2:36
M. NestorM. Nestor
778113
$endgroup$
$begingroup$
Isnt the upper limit x ?
$endgroup$
– user599310
Jan 16 at 3:37
$begingroup$
Yes, the lower bound is $y_min=x^{1/2}$ and the upper bound is $y_max=x$.
$endgroup$
– M. Nestor
Jan 16 at 3:44
$begingroup$
i am trying to do the calculations but are too hard
$endgroup$
– user599310
Jan 18 at 22:32
$begingroup$
I edited my answer, perhaps you can complete the calculation.
$endgroup$
– M. Nestor
Jan 18 at 23:56
add a comment |
$begingroup$
We have a function in two variables $f(x,y)=x^2+y^2$ over the region bounded by the curves $x=1$, $x=2$, $y=x$, and $y=x^{1/2}$. Here is a Desmos plot of these curves:
The only region bounded by all four curves seems to be the curved triangle containing $(1.5,1.5)$. To set up the integral find lower and upper bounds on both variables. It is $1leq xleq2$ and $x^{1/2}<y<x$. Then we have the double integral:
$$I=int_{x=1}^2int_{y=x^{1/2}}^x x^2+y^2text{ }dytext{ }dx$$
First take the antiderivative of $f(x,y)$ with respect to $y$:
$$=int_{x=1}^2Big[x^2y+frac{1}{3}y^3Big]_{y=x^{1/2}}^xdx$$
Evaluate from $y=x^{1/2}$ to $x$:
$$=int_{x=1}^2Big[Big(x^3+frac{1}{3}x^3Big)-Big(x^{5/2}+frac{1}{3}x^{3/2}Big)Big]dx$$
$$=int_{x=1}^2frac{4}{3}x^3-x^{5/2}-frac{1}{3}x^{3/2}text{ }dx$$
answered Jan 16 at 2:36
M. NestorM. Nestor
778113
$endgroup$
$begingroup$
Isnt the upper limit x ?
$endgroup$
– user599310
Jan 16 at 3:37
$begingroup$
Yes, the lower bound is $y_min=x^{1/2}$ and the upper bound is $y_max=x$.
$endgroup$
– M. Nestor
Jan 16 at 3:44
$begingroup$
i am trying to do the calculations but are too hard
$endgroup$
– user599310
Jan 18 at 22:32
$begingroup$
I edited my answer, perhaps you can complete the calculation.
$endgroup$
– M. Nestor
Jan 18 at 23:56
add a comment |
$begingroup$
We have a function in two variables $f(x,y)=x^2+y^2$ over the region bounded by the curves $x=1$, $x=2$, $y=x$, and $y=x^{1/2}$. Here is a Desmos plot of these curves:
The only region bounded by all four curves seems to be the curved triangle containing $(1.5,1.5)$. To set up the integral find lower and upper bounds on both variables. It is $1leq xleq2$ and $x^{1/2}<y<x$. Then we have the double integral:
$$I=int_{x=1}^2int_{y=x^{1/2}}^x x^2+y^2text{ }dytext{ }dx$$
First take the antiderivative of $f(x,y)$ with respect to $y$:
$$=int_{x=1}^2Big[x^2y+frac{1}{3}y^3Big]_{y=x^{1/2}}^xdx$$
Evaluate from $y=x^{1/2}$ to $x$:
$$=int_{x=1}^2Big[Big(x^3+frac{1}{3}x^3Big)-Big(x^{5/2}+frac{1}{3}x^{3/2}Big)Big]dx$$
$$=int_{x=1}^2frac{4}{3}x^3-x^{5/2}-frac{1}{3}x^{3/2}text{ }dx$$
answered Jan 16 at 2:36
M. NestorM. Nestor
778113
$endgroup$
We have a function in two variables $f(x,y)=x^2+y^2$ over the region bounded by the curves $x=1$, $x=2$, $y=x$, and $y=x^{1/2}$. Here is a Desmos plot of these curves:
The only region bounded by all four curves seems to be the curved triangle containing $(1.5,1.5)$. To set up the integral find lower and upper bounds on both variables. It is $1leq xleq2$ and $x^{1/2}<y<x$. Then we have the double integral:
$$I=int_{x=1}^2int_{y=x^{1/2}}^x x^2+y^2text{ }dytext{ }dx$$
First take the antiderivative of $f(x,y)$ with respect to $y$:
$$=int_{x=1}^2Big[x^2y+frac{1}{3}y^3Big]_{y=x^{1/2}}^xdx$$
Evaluate from $y=x^{1/2}$ to $x$:
$$=int_{x=1}^2Big[Big(x^3+frac{1}{3}x^3Big)-Big(x^{5/2}+frac{1}{3}x^{3/2}Big)Big]dx$$
$$=int_{x=1}^2frac{4}{3}x^3-x^{5/2}-frac{1}{3}x^{3/2}text{ }dx$$
answered Jan 16 at 2:36
M. NestorM. Nestor
778113
edited Jan 18 at 23:56
answered Jan 16 at 2:36
M. NestorM. Nestor
778113
answered Jan 16 at 2:36
M. NestorM. Nestor
778113
answered Jan 16 at 2:36
M. NestorM. Nestor
778113
778113
$begingroup$
Isnt the upper limit x ?
$endgroup$
– user599310
Jan 16 at 3:37
$begingroup$
Yes, the lower bound is $y_min=x^{1/2}$ and the upper bound is $y_max=x$.
$endgroup$
– M. Nestor
Jan 16 at 3:44
$begingroup$
i am trying to do the calculations but are too hard
$endgroup$
– user599310
Jan 18 at 22:32
$begingroup$
I edited my answer, perhaps you can complete the calculation.
$endgroup$
– M. Nestor
Jan 18 at 23:56
add a comment |
$begingroup$
Isnt the upper limit x ?
$endgroup$
– user599310
Jan 16 at 3:37
$begingroup$
Yes, the lower bound is $y_min=x^{1/2}$ and the upper bound is $y_max=x$.
$endgroup$
– M. Nestor
Jan 16 at 3:44
$begingroup$
i am trying to do the calculations but are too hard
$endgroup$
– user599310
Jan 18 at 22:32
$begingroup$
I edited my answer, perhaps you can complete the calculation.
$endgroup$
– M. Nestor
Jan 18 at 23:56
$begingroup$
Isnt the upper limit x ?
$endgroup$
– user599310
Jan 16 at 3:37
$begingroup$
Isnt the upper limit x ?
$endgroup$
– user599310
Jan 16 at 3:37
$begingroup$
Yes, the lower bound is $y_min=x^{1/2}$ and the upper bound is $y_max=x$.
$endgroup$
– M. Nestor
Jan 16 at 3:44
$begingroup$
Yes, the lower bound is $y_min=x^{1/2}$ and the upper bound is $y_max=x$.
$endgroup$
– M. Nestor
Jan 16 at 3:44
$begingroup$
i am trying to do the calculations but are too hard
$endgroup$
– user599310
Jan 18 at 22:32
$begingroup$
i am trying to do the calculations but are too hard
$endgroup$
– user599310
Jan 18 at 22:32
$begingroup$
I edited my answer, perhaps you can complete the calculation.
$endgroup$
– M. Nestor
Jan 18 at 23:56
$begingroup$
I edited my answer, perhaps you can complete the calculation.
$endgroup$
– M. Nestor
Jan 18 at 23:56
add a comment |
