Adding double and triple integrals
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As I know, we can easily add single integrals as follows: $$int_a^bfdx + int_b^cfdx = int_a^cfdx $$
However, I'm not sure how this works for double and triple integrals? For example, suppose I have a discontinuous volume that I want to find with the following bounds:
- z: $x^2 to 3-y$
- x: $-sqrt3 to -1$ and $1 to sqrt3$
- y: $0 to 2$
Can I write this as the triple integral:
$$int_0^2 int_{-sqrt3}^{-1} int_{x^2}^{3-y} f dzdxdy + int_0^2 int_{1}^{sqrt3} int_{x^2}^{3-y} f dzdxdy $$
integration definite-integrals
asked Dec 3 '18 at 9:27
Gummy bearsGummy bears
1,88311531
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add a comment |
$begingroup$
As I know, we can easily add single integrals as follows: $$int_a^bfdx + int_b^cfdx = int_a^cfdx $$
However, I'm not sure how this works for double and triple integrals? For example, suppose I have a discontinuous volume that I want to find with the following bounds:
- z: $x^2 to 3-y$
- x: $-sqrt3 to -1$ and $1 to sqrt3$
- y: $0 to 2$
Can I write this as the triple integral:
$$int_0^2 int_{-sqrt3}^{-1} int_{x^2}^{3-y} f dzdxdy + int_0^2 int_{1}^{sqrt3} int_{x^2}^{3-y} f dzdxdy $$
integration definite-integrals
asked Dec 3 '18 at 9:27
Gummy bearsGummy bears
1,88311531
$endgroup$
$begingroup$
You surely can. Integrals over disjoint regions add up.
$endgroup$
– Kavi Rama Murthy
Dec 3 '18 at 9:29
add a comment |
$begingroup$
As I know, we can easily add single integrals as follows: $$int_a^bfdx + int_b^cfdx = int_a^cfdx $$
However, I'm not sure how this works for double and triple integrals? For example, suppose I have a discontinuous volume that I want to find with the following bounds:
- z: $x^2 to 3-y$
- x: $-sqrt3 to -1$ and $1 to sqrt3$
- y: $0 to 2$
Can I write this as the triple integral:
$$int_0^2 int_{-sqrt3}^{-1} int_{x^2}^{3-y} f dzdxdy + int_0^2 int_{1}^{sqrt3} int_{x^2}^{3-y} f dzdxdy $$
integration definite-integrals
asked Dec 3 '18 at 9:27
Gummy bearsGummy bears
1,88311531
$endgroup$
As I know, we can easily add single integrals as follows: $$int_a^bfdx + int_b^cfdx = int_a^cfdx $$
However, I'm not sure how this works for double and triple integrals? For example, suppose I have a discontinuous volume that I want to find with the following bounds:
- z: $x^2 to 3-y$
- x: $-sqrt3 to -1$ and $1 to sqrt3$
- y: $0 to 2$
Can I write this as the triple integral:
$$int_0^2 int_{-sqrt3}^{-1} int_{x^2}^{3-y} f dzdxdy + int_0^2 int_{1}^{sqrt3} int_{x^2}^{3-y} f dzdxdy $$
integration definite-integrals
integration definite-integrals
asked Dec 3 '18 at 9:27
Gummy bearsGummy bears
1,88311531
asked Dec 3 '18 at 9:27
Gummy bearsGummy bears
1,88311531
asked Dec 3 '18 at 9:27
Gummy bearsGummy bears
1,88311531
asked Dec 3 '18 at 9:27
Gummy bearsGummy bears
1,88311531
asked Dec 3 '18 at 9:27
Gummy bearsGummy bears
1,88311531
1,88311531
$begingroup$
You surely can. Integrals over disjoint regions add up.
$endgroup$
– Kavi Rama Murthy
Dec 3 '18 at 9:29
add a comment |
$begingroup$
You surely can. Integrals over disjoint regions add up.
$endgroup$
– Kavi Rama Murthy
Dec 3 '18 at 9:29
$begingroup$
You surely can. Integrals over disjoint regions add up.
$endgroup$
– Kavi Rama Murthy
Dec 3 '18 at 9:29
$begingroup$
You surely can. Integrals over disjoint regions add up.
$endgroup$
– Kavi Rama Murthy
Dec 3 '18 at 9:29
add a comment |
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$begingroup$
You surely can. Integrals over disjoint regions add up.
$endgroup$
– Kavi Rama Murthy
Dec 3 '18 at 9:29