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A conversion of double integral to polar and evaluate, check
How to compute this integral by converting it into polar coordinate?Question concerning the domain of polar coordinate.Why am I evaluating this polar integral wrong?Conversion of an integral in cartesian to polar coordinatesWhere am I going Wrong in this Polar Coordinate Conversion?Double integral with Polar coordinates - hard exampleConvert the integral from rectangular to cylindrical coordinates and solveEvaluating integral by changing to polar coords $int int x^2y da$Double Integral Polar ConversionDouble integral conversions to Polar
$begingroup$
Problem 1
Convert to polar form and solve
$$int^1_0int_0^sqrt2y-y^2(1-x^2-y^2)text dx dy$$
$x^2+y^2-2y=0$
$x^2+(y-1)^2=1$
$x=rcostheta$
$y=rsintheta+1$
$r^2=1, r=1$
$$int^pi_0int^1_0(1-r^2cos^2theta-(y^2cos^2theta+1)) rdrdtheta$$
$$int^pi_0int^1_0(-r^3)drdtheta = -1/4pi$$
calculus integration multivariable-calculus proof-verification polar-coordinates
asked Mar 17 at 20:51
MasterYoshiMasterYoshi
807
$endgroup$
add a comment |
$begingroup$
Problem 1
Convert to polar form and solve
$$int^1_0int_0^sqrt2y-y^2(1-x^2-y^2)text dx dy$$
$x^2+y^2-2y=0$
$x^2+(y-1)^2=1$
$x=rcostheta$
$y=rsintheta+1$
$r^2=1, r=1$
$$int^pi_0int^1_0(1-r^2cos^2theta-(y^2cos^2theta+1)) rdrdtheta$$
$$int^pi_0int^1_0(-r^3)drdtheta = -1/4pi$$
calculus integration multivariable-calculus proof-verification polar-coordinates
asked Mar 17 at 20:51
MasterYoshiMasterYoshi
807
$endgroup$
1
$begingroup$
What is the question?
$endgroup$
– MisterRiemann
Mar 17 at 20:52
$begingroup$
is this correct? i do not have the answer to this problem
$endgroup$
– MasterYoshi
Mar 17 at 20:54
add a comment |
$begingroup$
Problem 1
Convert to polar form and solve
$$int^1_0int_0^sqrt2y-y^2(1-x^2-y^2)text dx dy$$
$x^2+y^2-2y=0$
$x^2+(y-1)^2=1$
$x=rcostheta$
$y=rsintheta+1$
$r^2=1, r=1$
$$int^pi_0int^1_0(1-r^2cos^2theta-(y^2cos^2theta+1)) rdrdtheta$$
$$int^pi_0int^1_0(-r^3)drdtheta = -1/4pi$$
calculus integration multivariable-calculus proof-verification polar-coordinates
asked Mar 17 at 20:51
MasterYoshiMasterYoshi
807
$endgroup$
Problem 1
Convert to polar form and solve
$$int^1_0int_0^sqrt2y-y^2(1-x^2-y^2)text dx dy$$
$x^2+y^2-2y=0$
$x^2+(y-1)^2=1$
$x=rcostheta$
$y=rsintheta+1$
$r^2=1, r=1$
$$int^pi_0int^1_0(1-r^2cos^2theta-(y^2cos^2theta+1)) rdrdtheta$$
$$int^pi_0int^1_0(-r^3)drdtheta = -1/4pi$$
calculus integration multivariable-calculus proof-verification polar-coordinates
calculus integration multivariable-calculus proof-verification polar-coordinates
asked Mar 17 at 20:51
MasterYoshiMasterYoshi
807
asked Mar 17 at 20:51
MasterYoshiMasterYoshi
807
asked Mar 17 at 20:51
MasterYoshiMasterYoshi
807
asked Mar 17 at 20:51
MasterYoshiMasterYoshi
807
asked Mar 17 at 20:51
MasterYoshiMasterYoshi
807
807
1
$begingroup$
What is the question?
$endgroup$
– MisterRiemann
Mar 17 at 20:52
$begingroup$
is this correct? i do not have the answer to this problem
$endgroup$
– MasterYoshi
Mar 17 at 20:54
add a comment |
1
$begingroup$
What is the question?
$endgroup$
– MisterRiemann
Mar 17 at 20:52
$begingroup$
is this correct? i do not have the answer to this problem
$endgroup$
– MasterYoshi
Mar 17 at 20:54
1
1
$begingroup$
What is the question?
$endgroup$
– MisterRiemann
Mar 17 at 20:52
$begingroup$
What is the question?
$endgroup$
– MisterRiemann
Mar 17 at 20:52
$begingroup$
is this correct? i do not have the answer to this problem
$endgroup$
– MasterYoshi
Mar 17 at 20:54
$begingroup$
is this correct? i do not have the answer to this problem
$endgroup$
– MasterYoshi
Mar 17 at 20:54
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You made an error when you substituted $y^2$. For the reference, the correct result is $frac23 - fracpi8$.
answered Mar 17 at 21:04
KlausKlaus
2,782113
$endgroup$
$begingroup$
Hey, i dont know why i cannot see that mistake? you mean when i plugged $ y= rsintheta+1$ in $y^2$?
$endgroup$
– MasterYoshi
Mar 17 at 21:13
$begingroup$
$y^2 = (rsintheta+1)^2$ and not what you did in the OP (if I'm reading this correctly).
$endgroup$
– Klaus
Mar 17 at 21:19
$begingroup$
ohh, so $(1-r^2costheta-(r^2sin^2theta+2rsintheta+1))$? i get the integral of $-r^3-2r^2sintheta$ ?
$endgroup$
– MasterYoshi
Mar 17 at 21:44
$begingroup$
the result i got from the above is$-4/3 -pi/4$ so the plugging is still incorrect?
$endgroup$
– MasterYoshi
Mar 17 at 22:16
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You made an error when you substituted $y^2$. For the reference, the correct result is $frac23 - fracpi8$.
answered Mar 17 at 21:04
KlausKlaus
2,782113
$endgroup$
$begingroup$
Hey, i dont know why i cannot see that mistake? you mean when i plugged $ y= rsintheta+1$ in $y^2$?
$endgroup$
– MasterYoshi
Mar 17 at 21:13
$begingroup$
$y^2 = (rsintheta+1)^2$ and not what you did in the OP (if I'm reading this correctly).
$endgroup$
– Klaus
Mar 17 at 21:19
$begingroup$
ohh, so $(1-r^2costheta-(r^2sin^2theta+2rsintheta+1))$? i get the integral of $-r^3-2r^2sintheta$ ?
$endgroup$
– MasterYoshi
Mar 17 at 21:44
$begingroup$
the result i got from the above is$-4/3 -pi/4$ so the plugging is still incorrect?
$endgroup$
– MasterYoshi
Mar 17 at 22:16
add a comment |
$begingroup$
You made an error when you substituted $y^2$. For the reference, the correct result is $frac23 - fracpi8$.
answered Mar 17 at 21:04
KlausKlaus
2,782113
$endgroup$
$begingroup$
Hey, i dont know why i cannot see that mistake? you mean when i plugged $ y= rsintheta+1$ in $y^2$?
$endgroup$
– MasterYoshi
Mar 17 at 21:13
$begingroup$
$y^2 = (rsintheta+1)^2$ and not what you did in the OP (if I'm reading this correctly).
$endgroup$
– Klaus
Mar 17 at 21:19
$begingroup$
ohh, so $(1-r^2costheta-(r^2sin^2theta+2rsintheta+1))$? i get the integral of $-r^3-2r^2sintheta$ ?
$endgroup$
– MasterYoshi
Mar 17 at 21:44
$begingroup$
the result i got from the above is$-4/3 -pi/4$ so the plugging is still incorrect?
$endgroup$
– MasterYoshi
Mar 17 at 22:16
add a comment |
$begingroup$
You made an error when you substituted $y^2$. For the reference, the correct result is $frac23 - fracpi8$.
answered Mar 17 at 21:04
KlausKlaus
2,782113
$endgroup$
You made an error when you substituted $y^2$. For the reference, the correct result is $frac23 - fracpi8$.
answered Mar 17 at 21:04
KlausKlaus
2,782113
answered Mar 17 at 21:04
KlausKlaus
2,782113
answered Mar 17 at 21:04
KlausKlaus
2,782113
answered Mar 17 at 21:04
KlausKlaus
2,782113
2,782113
$begingroup$
Hey, i dont know why i cannot see that mistake? you mean when i plugged $ y= rsintheta+1$ in $y^2$?
$endgroup$
– MasterYoshi
Mar 17 at 21:13
$begingroup$
$y^2 = (rsintheta+1)^2$ and not what you did in the OP (if I'm reading this correctly).
$endgroup$
– Klaus
Mar 17 at 21:19
$begingroup$
ohh, so $(1-r^2costheta-(r^2sin^2theta+2rsintheta+1))$? i get the integral of $-r^3-2r^2sintheta$ ?
$endgroup$
– MasterYoshi
Mar 17 at 21:44
$begingroup$
the result i got from the above is$-4/3 -pi/4$ so the plugging is still incorrect?
$endgroup$
– MasterYoshi
Mar 17 at 22:16
add a comment |
$begingroup$
Hey, i dont know why i cannot see that mistake? you mean when i plugged $ y= rsintheta+1$ in $y^2$?
$endgroup$
– MasterYoshi
Mar 17 at 21:13
$begingroup$
$y^2 = (rsintheta+1)^2$ and not what you did in the OP (if I'm reading this correctly).
$endgroup$
– Klaus
Mar 17 at 21:19
$begingroup$
ohh, so $(1-r^2costheta-(r^2sin^2theta+2rsintheta+1))$? i get the integral of $-r^3-2r^2sintheta$ ?
$endgroup$
– MasterYoshi
Mar 17 at 21:44
$begingroup$
the result i got from the above is$-4/3 -pi/4$ so the plugging is still incorrect?
$endgroup$
– MasterYoshi
Mar 17 at 22:16
$begingroup$
Hey, i dont know why i cannot see that mistake? you mean when i plugged $ y= rsintheta+1$ in $y^2$?
$endgroup$
– MasterYoshi
Mar 17 at 21:13
$begingroup$
Hey, i dont know why i cannot see that mistake? you mean when i plugged $ y= rsintheta+1$ in $y^2$?
$endgroup$
– MasterYoshi
Mar 17 at 21:13
$begingroup$
$y^2 = (rsintheta+1)^2$ and not what you did in the OP (if I'm reading this correctly).
$endgroup$
– Klaus
Mar 17 at 21:19
$begingroup$
$y^2 = (rsintheta+1)^2$ and not what you did in the OP (if I'm reading this correctly).
$endgroup$
– Klaus
Mar 17 at 21:19
$begingroup$
ohh, so $(1-r^2costheta-(r^2sin^2theta+2rsintheta+1))$? i get the integral of $-r^3-2r^2sintheta$ ?
$endgroup$
– MasterYoshi
Mar 17 at 21:44
$begingroup$
ohh, so $(1-r^2costheta-(r^2sin^2theta+2rsintheta+1))$? i get the integral of $-r^3-2r^2sintheta$ ?
$endgroup$
– MasterYoshi
Mar 17 at 21:44
$begingroup$
the result i got from the above is$-4/3 -pi/4$ so the plugging is still incorrect?
$endgroup$
– MasterYoshi
Mar 17 at 22:16
$begingroup$
the result i got from the above is$-4/3 -pi/4$ so the plugging is still incorrect?
$endgroup$
– MasterYoshi
Mar 17 at 22:16
add a comment |
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1
$begingroup$
What is the question?
$endgroup$
– MisterRiemann
Mar 17 at 20:52
$begingroup$
is this correct? i do not have the answer to this problem
$endgroup$
– MasterYoshi
Mar 17 at 20:54