How to integrate $ ycos^4x$ over a disk?
$begingroup$
I'm trying to integrate
$$int_D ycos^4x , mathrm dx mathrm dy$$
where $D={(x,y)in R^2;x^2+y^2<pi}.$
I'm thinking using the polar coordinates but doing so the integral would become,
$$int_D rho sinthetacos^4(rho costheta) , mathrm dtheta mathrm drho$$
and $D={(rho,theta)in R^2;rho^2<pi,0leq theta leq 2pi}$ which is even more difficult. Can someone explain me how to solve the exercise?
The professor says that this should have a quick resolution, so probably there is something I'm missing, please help me
calculus integration analysis multivariable-calculus polar-coordinates
edited Jan 13 at 23:59
JamalS
735417
asked Jan 13 at 23:27
MarcoMarco
307
$endgroup$
add a comment |
$begingroup$
I'm trying to integrate
$$int_D ycos^4x , mathrm dx mathrm dy$$
where $D={(x,y)in R^2;x^2+y^2<pi}.$
I'm thinking using the polar coordinates but doing so the integral would become,
$$int_D rho sinthetacos^4(rho costheta) , mathrm dtheta mathrm drho$$
and $D={(rho,theta)in R^2;rho^2<pi,0leq theta leq 2pi}$ which is even more difficult. Can someone explain me how to solve the exercise?
The professor says that this should have a quick resolution, so probably there is something I'm missing, please help me
calculus integration analysis multivariable-calculus polar-coordinates
edited Jan 13 at 23:59
JamalS
735417
asked Jan 13 at 23:27
MarcoMarco
307
$endgroup$
1
$begingroup$
What happens if you write $(x,y)=(x’,-y’)$?
$endgroup$
– Mindlack
Jan 13 at 23:34
1
$begingroup$
The function is odd on $y$ , and the domain is symmetric with respect to $y$: this means that the integral is $0$.
$endgroup$
– Crostul
Jan 13 at 23:39
$begingroup$
@Crostul Ok this is exactly the quick resolution I mentioned, but I can use the odd property because D is centered in (0,0) , right?
$endgroup$
– Marco
Jan 14 at 9:20
add a comment |
$begingroup$
I'm trying to integrate
$$int_D ycos^4x , mathrm dx mathrm dy$$
where $D={(x,y)in R^2;x^2+y^2<pi}.$
I'm thinking using the polar coordinates but doing so the integral would become,
$$int_D rho sinthetacos^4(rho costheta) , mathrm dtheta mathrm drho$$
and $D={(rho,theta)in R^2;rho^2<pi,0leq theta leq 2pi}$ which is even more difficult. Can someone explain me how to solve the exercise?
The professor says that this should have a quick resolution, so probably there is something I'm missing, please help me
calculus integration analysis multivariable-calculus polar-coordinates
edited Jan 13 at 23:59
JamalS
735417
asked Jan 13 at 23:27
MarcoMarco
307
$endgroup$
I'm trying to integrate
$$int_D ycos^4x , mathrm dx mathrm dy$$
where $D={(x,y)in R^2;x^2+y^2<pi}.$
I'm thinking using the polar coordinates but doing so the integral would become,
$$int_D rho sinthetacos^4(rho costheta) , mathrm dtheta mathrm drho$$
and $D={(rho,theta)in R^2;rho^2<pi,0leq theta leq 2pi}$ which is even more difficult. Can someone explain me how to solve the exercise?
The professor says that this should have a quick resolution, so probably there is something I'm missing, please help me
calculus integration analysis multivariable-calculus polar-coordinates
calculus integration analysis multivariable-calculus polar-coordinates
edited Jan 13 at 23:59
JamalS
735417
asked Jan 13 at 23:27
MarcoMarco
307
edited Jan 13 at 23:59
JamalS
735417
asked Jan 13 at 23:27
MarcoMarco
307
edited Jan 13 at 23:59
JamalS
735417
edited Jan 13 at 23:59
JamalS
735417
edited Jan 13 at 23:59
JamalS
735417
735417
asked Jan 13 at 23:27
MarcoMarco
307
asked Jan 13 at 23:27
MarcoMarco
307
asked Jan 13 at 23:27
MarcoMarco
307
307
1
$begingroup$
What happens if you write $(x,y)=(x’,-y’)$?
$endgroup$
– Mindlack
Jan 13 at 23:34
1
$begingroup$
The function is odd on $y$ , and the domain is symmetric with respect to $y$: this means that the integral is $0$.
$endgroup$
– Crostul
Jan 13 at 23:39
$begingroup$
@Crostul Ok this is exactly the quick resolution I mentioned, but I can use the odd property because D is centered in (0,0) , right?
$endgroup$
– Marco
Jan 14 at 9:20
add a comment |
1
$begingroup$
What happens if you write $(x,y)=(x’,-y’)$?
$endgroup$
– Mindlack
Jan 13 at 23:34
1
$begingroup$
The function is odd on $y$ , and the domain is symmetric with respect to $y$: this means that the integral is $0$.
$endgroup$
– Crostul
Jan 13 at 23:39
$begingroup$
@Crostul Ok this is exactly the quick resolution I mentioned, but I can use the odd property because D is centered in (0,0) , right?
$endgroup$
– Marco
Jan 14 at 9:20
1
1
$begingroup$
What happens if you write $(x,y)=(x’,-y’)$?
$endgroup$
– Mindlack
Jan 13 at 23:34
$begingroup$
What happens if you write $(x,y)=(x’,-y’)$?
$endgroup$
– Mindlack
Jan 13 at 23:34
1
1
$begingroup$
The function is odd on $y$ , and the domain is symmetric with respect to $y$: this means that the integral is $0$.
$endgroup$
– Crostul
Jan 13 at 23:39
$begingroup$
The function is odd on $y$ , and the domain is symmetric with respect to $y$: this means that the integral is $0$.
$endgroup$
– Crostul
Jan 13 at 23:39
$begingroup$
@Crostul Ok this is exactly the quick resolution I mentioned, but I can use the odd property because D is centered in (0,0) , right?
$endgroup$
– Marco
Jan 14 at 9:20
$begingroup$
@Crostul Ok this is exactly the quick resolution I mentioned, but I can use the odd property because D is centered in (0,0) , right?
$endgroup$
– Marco
Jan 14 at 9:20
add a comment |
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1
$begingroup$
What happens if you write $(x,y)=(x’,-y’)$?
$endgroup$
– Mindlack
Jan 13 at 23:34
1
$begingroup$
The function is odd on $y$ , and the domain is symmetric with respect to $y$: this means that the integral is $0$.
$endgroup$
– Crostul
Jan 13 at 23:39
$begingroup$
@Crostul Ok this is exactly the quick resolution I mentioned, but I can use the odd property because D is centered in (0,0) , right?
$endgroup$
– Marco
Jan 14 at 9:20