evaluate the surface integral $f(x, y, z) = x^2 + y^2$ and $S$ is part of the cone $z = sqrtx^2 + y^2$ with $z leq 2$Evaluate Surface IntegralSurface Integral of a Right Circular ConeEvaluate the surface integral from the paraboloidSurface integral with domain that contains an infinite coneEvaluate $iint dydx$ on the domain $0leq rleq1$, $pi/3leqtheta leq2pi/3$How to calculate the surface area of a cone using cylindrical coordinates?Calculating a surface integral between a cone and a hemisphereSurface integral - cone below planeVolume between cone and sphere of radius $sqrt2$ with surface integral
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evaluate the surface integral $f(x, y, z) = x^2 + y^2$ and $S$ is part of the cone $z = sqrtx^2 + y^2$ with $z leq 2$
Evaluate Surface IntegralSurface Integral of a Right Circular ConeEvaluate the surface integral from the paraboloidSurface integral with domain that contains an infinite coneEvaluate $iint dydx$ on the domain $0leq rleq1$, $pi/3leqtheta leq2pi/3$How to calculate the surface area of a cone using cylindrical coordinates?Calculating a surface integral between a cone and a hemisphereSurface integral - cone below planeVolume between cone and sphere of radius $sqrt2$ with surface integral
$begingroup$
Evaluate the surface integral of scalar function $int_S f dS$
$f(x, y, z) = x^2 + y^2$ and $S$ is part of the cone $z = sqrtx^2 + y^2$ with $z leq 2$
my attempt
$f(x,y,z) = x^2 + y^2$ and $z = sqrtx^2 + y^2, z leq 2$
$fracdzdx = fracxsqrtx^2 + y^2$ and $fracdzdy = fracysqrtx^2 + y^2$
$ds = sqrtleft(fracdzdx right)^2 + left(fracdzdy right)^2 dx dy= sqrtfracx^2x^2 + y^2 + fracy^2x^2 + y^2dx dy=dx dy$
$int_S f dS = iint(x^2 + y^2)dx dy$, such that $x^2 + y^2 = 4, z leq 2$
$int_0^2pi int_0^2 r^2 r dr d theta = 8 pi$
would this be right?
multivariable-calculus
edited Mar 4 at 8:15
Robert Z
100k1069141
asked Mar 4 at 6:52
tomtom
184
$endgroup$
add a comment |
$begingroup$
Evaluate the surface integral of scalar function $int_S f dS$
$f(x, y, z) = x^2 + y^2$ and $S$ is part of the cone $z = sqrtx^2 + y^2$ with $z leq 2$
my attempt
$f(x,y,z) = x^2 + y^2$ and $z = sqrtx^2 + y^2, z leq 2$
$fracdzdx = fracxsqrtx^2 + y^2$ and $fracdzdy = fracysqrtx^2 + y^2$
$ds = sqrtleft(fracdzdx right)^2 + left(fracdzdy right)^2 dx dy= sqrtfracx^2x^2 + y^2 + fracy^2x^2 + y^2dx dy=dx dy$
$int_S f dS = iint(x^2 + y^2)dx dy$, such that $x^2 + y^2 = 4, z leq 2$
$int_0^2pi int_0^2 r^2 r dr d theta = 8 pi$
would this be right?
multivariable-calculus
edited Mar 4 at 8:15
Robert Z
100k1069141
asked Mar 4 at 6:52
tomtom
184
$endgroup$
$begingroup$
Yes, it is right, but for a small typo where your write "such that $x^2+y^2=4$" , where it should be $x^2+y^2leq 4$.
$endgroup$
– uniquesolution
Mar 4 at 8:34
add a comment |
$begingroup$
Evaluate the surface integral of scalar function $int_S f dS$
$f(x, y, z) = x^2 + y^2$ and $S$ is part of the cone $z = sqrtx^2 + y^2$ with $z leq 2$
my attempt
$f(x,y,z) = x^2 + y^2$ and $z = sqrtx^2 + y^2, z leq 2$
$fracdzdx = fracxsqrtx^2 + y^2$ and $fracdzdy = fracysqrtx^2 + y^2$
$ds = sqrtleft(fracdzdx right)^2 + left(fracdzdy right)^2 dx dy= sqrtfracx^2x^2 + y^2 + fracy^2x^2 + y^2dx dy=dx dy$
$int_S f dS = iint(x^2 + y^2)dx dy$, such that $x^2 + y^2 = 4, z leq 2$
$int_0^2pi int_0^2 r^2 r dr d theta = 8 pi$
would this be right?
multivariable-calculus
edited Mar 4 at 8:15
Robert Z
100k1069141
asked Mar 4 at 6:52
tomtom
184
$endgroup$
Evaluate the surface integral of scalar function $int_S f dS$
$f(x, y, z) = x^2 + y^2$ and $S$ is part of the cone $z = sqrtx^2 + y^2$ with $z leq 2$
my attempt
$f(x,y,z) = x^2 + y^2$ and $z = sqrtx^2 + y^2, z leq 2$
$fracdzdx = fracxsqrtx^2 + y^2$ and $fracdzdy = fracysqrtx^2 + y^2$
$ds = sqrtleft(fracdzdx right)^2 + left(fracdzdy right)^2 dx dy= sqrtfracx^2x^2 + y^2 + fracy^2x^2 + y^2dx dy=dx dy$
$int_S f dS = iint(x^2 + y^2)dx dy$, such that $x^2 + y^2 = 4, z leq 2$
$int_0^2pi int_0^2 r^2 r dr d theta = 8 pi$
would this be right?
multivariable-calculus
multivariable-calculus
edited Mar 4 at 8:15
Robert Z
100k1069141
asked Mar 4 at 6:52
tomtom
184
edited Mar 4 at 8:15
Robert Z
100k1069141
asked Mar 4 at 6:52
tomtom
184
edited Mar 4 at 8:15
Robert Z
100k1069141
edited Mar 4 at 8:15
Robert Z
100k1069141
edited Mar 4 at 8:15
Robert Z
100k1069141
100k1069141
asked Mar 4 at 6:52
tomtom
184
asked Mar 4 at 6:52
tomtom
184
asked Mar 4 at 6:52
tomtom
184
184
$begingroup$
Yes, it is right, but for a small typo where your write "such that $x^2+y^2=4$" , where it should be $x^2+y^2leq 4$.
$endgroup$
– uniquesolution
Mar 4 at 8:34
add a comment |
$begingroup$
Yes, it is right, but for a small typo where your write "such that $x^2+y^2=4$" , where it should be $x^2+y^2leq 4$.
$endgroup$
– uniquesolution
Mar 4 at 8:34
$begingroup$
Yes, it is right, but for a small typo where your write "such that $x^2+y^2=4$" , where it should be $x^2+y^2leq 4$.
$endgroup$
– uniquesolution
Mar 4 at 8:34
$begingroup$
Yes, it is right, but for a small typo where your write "such that $x^2+y^2=4$" , where it should be $x^2+y^2leq 4$.
$endgroup$
– uniquesolution
Mar 4 at 8:34
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your approach is almost correct but for one erroneous formula:
$mathrmdS = sqrtz^2_x+z^2_y + 1 mathrmdA $
$$
beginalign*
mathrmdvecS = langle -z_x,-z_y, 1 rangle mathrmdA
endalign*
$$
therefore, $$|mathrmdvecS| = sqrtz^2_x+z^2_y + 1 mathrmdA $$
$z_x$ and $z_y$ values what you got are correct.
Let's proceed:
$$
beginalign*
iint_S fmathrmdS &= iint_S x^2+y^2 sqrtz^2_x+z^2_y + 1 mathrmdA \
\
&= iint_S x^2+y^2 sqrtfracx^2x^2+y^2+fracy^2x^2+y^2+1 mathrmdxmathrmdy \
\
endalign*
$$
Now switch to polar coordinates:
$$
beginalign*
iint_S fmathrmdS &= iint_S r^2sqrt2 cdot rmathrmdrmathrmdrtheta \
\
&= sqrt2int_0^2piint_0^2r^3 mathrmdrtheta \
\
&= 8sqrt2pi
endalign*
$$
answered Mar 10 at 19:01
DubsVeer23DubsVeer23
682
$endgroup$
add a comment |
Your Answer
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your approach is almost correct but for one erroneous formula:
$mathrmdS = sqrtz^2_x+z^2_y + 1 mathrmdA $
$$
beginalign*
mathrmdvecS = langle -z_x,-z_y, 1 rangle mathrmdA
endalign*
$$
therefore, $$|mathrmdvecS| = sqrtz^2_x+z^2_y + 1 mathrmdA $$
$z_x$ and $z_y$ values what you got are correct.
Let's proceed:
$$
beginalign*
iint_S fmathrmdS &= iint_S x^2+y^2 sqrtz^2_x+z^2_y + 1 mathrmdA \
\
&= iint_S x^2+y^2 sqrtfracx^2x^2+y^2+fracy^2x^2+y^2+1 mathrmdxmathrmdy \
\
endalign*
$$
Now switch to polar coordinates:
$$
beginalign*
iint_S fmathrmdS &= iint_S r^2sqrt2 cdot rmathrmdrmathrmdrtheta \
\
&= sqrt2int_0^2piint_0^2r^3 mathrmdrtheta \
\
&= 8sqrt2pi
endalign*
$$
answered Mar 10 at 19:01
DubsVeer23DubsVeer23
682
$endgroup$
add a comment |
$begingroup$
Your approach is almost correct but for one erroneous formula:
$mathrmdS = sqrtz^2_x+z^2_y + 1 mathrmdA $
$$
beginalign*
mathrmdvecS = langle -z_x,-z_y, 1 rangle mathrmdA
endalign*
$$
therefore, $$|mathrmdvecS| = sqrtz^2_x+z^2_y + 1 mathrmdA $$
$z_x$ and $z_y$ values what you got are correct.
Let's proceed:
$$
beginalign*
iint_S fmathrmdS &= iint_S x^2+y^2 sqrtz^2_x+z^2_y + 1 mathrmdA \
\
&= iint_S x^2+y^2 sqrtfracx^2x^2+y^2+fracy^2x^2+y^2+1 mathrmdxmathrmdy \
\
endalign*
$$
Now switch to polar coordinates:
$$
beginalign*
iint_S fmathrmdS &= iint_S r^2sqrt2 cdot rmathrmdrmathrmdrtheta \
\
&= sqrt2int_0^2piint_0^2r^3 mathrmdrtheta \
\
&= 8sqrt2pi
endalign*
$$
answered Mar 10 at 19:01
DubsVeer23DubsVeer23
682
$endgroup$
add a comment |
$begingroup$
Your approach is almost correct but for one erroneous formula:
$mathrmdS = sqrtz^2_x+z^2_y + 1 mathrmdA $
$$
beginalign*
mathrmdvecS = langle -z_x,-z_y, 1 rangle mathrmdA
endalign*
$$
therefore, $$|mathrmdvecS| = sqrtz^2_x+z^2_y + 1 mathrmdA $$
$z_x$ and $z_y$ values what you got are correct.
Let's proceed:
$$
beginalign*
iint_S fmathrmdS &= iint_S x^2+y^2 sqrtz^2_x+z^2_y + 1 mathrmdA \
\
&= iint_S x^2+y^2 sqrtfracx^2x^2+y^2+fracy^2x^2+y^2+1 mathrmdxmathrmdy \
\
endalign*
$$
Now switch to polar coordinates:
$$
beginalign*
iint_S fmathrmdS &= iint_S r^2sqrt2 cdot rmathrmdrmathrmdrtheta \
\
&= sqrt2int_0^2piint_0^2r^3 mathrmdrtheta \
\
&= 8sqrt2pi
endalign*
$$
answered Mar 10 at 19:01
DubsVeer23DubsVeer23
682
$endgroup$
Your approach is almost correct but for one erroneous formula:
$mathrmdS = sqrtz^2_x+z^2_y + 1 mathrmdA $
$$
beginalign*
mathrmdvecS = langle -z_x,-z_y, 1 rangle mathrmdA
endalign*
$$
therefore, $$|mathrmdvecS| = sqrtz^2_x+z^2_y + 1 mathrmdA $$
$z_x$ and $z_y$ values what you got are correct.
Let's proceed:
$$
beginalign*
iint_S fmathrmdS &= iint_S x^2+y^2 sqrtz^2_x+z^2_y + 1 mathrmdA \
\
&= iint_S x^2+y^2 sqrtfracx^2x^2+y^2+fracy^2x^2+y^2+1 mathrmdxmathrmdy \
\
endalign*
$$
Now switch to polar coordinates:
$$
beginalign*
iint_S fmathrmdS &= iint_S r^2sqrt2 cdot rmathrmdrmathrmdrtheta \
\
&= sqrt2int_0^2piint_0^2r^3 mathrmdrtheta \
\
&= 8sqrt2pi
endalign*
$$
answered Mar 10 at 19:01
DubsVeer23DubsVeer23
682
answered Mar 10 at 19:01
DubsVeer23DubsVeer23
682
answered Mar 10 at 19:01
DubsVeer23DubsVeer23
682
answered Mar 10 at 19:01
DubsVeer23DubsVeer23
682
682
add a comment |
add a comment |
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$begingroup$
Yes, it is right, but for a small typo where your write "such that $x^2+y^2=4$" , where it should be $x^2+y^2leq 4$.
$endgroup$
– uniquesolution
Mar 4 at 8:34