Mixing
Area of the surface
$begingroup$
Compute the area between the surfaces $(x^2+y^2)^2=a^2(x^2-y^2)$ and $x^2+y^2+z^2=a^2.$
My attempt:
I take the first case $zgeq 0$:
$$x=x$$
$$y=y$$
$$z=sqrt{a^2-(x^2+y^2)}$$
then
$$T_x=(1,0,frac {-x}{sqrt{a^2-(x^2+y^2)}})$$
$$T_x=(0,1,frac {-y}{sqrt{a^2-(x^2+y^2)}})$$
Then the area of the surface is:
$$A(S)=int_S dsigma=intint_D frac {a}{sqrt{a^2-(x^2+y^2)}}dxdy$$
By parameterization:
$$x=rcost$$
$$y=rsint$$
then $r=asqrt{cos(2t)}implies tin [0, frac {pi}4]$ and $rin[0,asqrt{cos(2t)}]$
$A(S)=(a^2-sqrt{2})$. But that's a wrong answer, so what am I doing wrong?
calculus integration area
edited Jan 28 at 7:14
BadAtGeometry
298215
asked Jan 28 at 7:08
C. CristiC. Cristi
1,639218
$endgroup$
add a comment |
$begingroup$
Compute the area between the surfaces $(x^2+y^2)^2=a^2(x^2-y^2)$ and $x^2+y^2+z^2=a^2.$
My attempt:
I take the first case $zgeq 0$:
$$x=x$$
$$y=y$$
$$z=sqrt{a^2-(x^2+y^2)}$$
then
$$T_x=(1,0,frac {-x}{sqrt{a^2-(x^2+y^2)}})$$
$$T_x=(0,1,frac {-y}{sqrt{a^2-(x^2+y^2)}})$$
Then the area of the surface is:
$$A(S)=int_S dsigma=intint_D frac {a}{sqrt{a^2-(x^2+y^2)}}dxdy$$
By parameterization:
$$x=rcost$$
$$y=rsint$$
then $r=asqrt{cos(2t)}implies tin [0, frac {pi}4]$ and $rin[0,asqrt{cos(2t)}]$
$A(S)=(a^2-sqrt{2})$. But that's a wrong answer, so what am I doing wrong?
calculus integration area
edited Jan 28 at 7:14
BadAtGeometry
298215
asked Jan 28 at 7:08
C. CristiC. Cristi
1,639218
$endgroup$
1
$begingroup$
It is unclear what you mean by the "area between the surfaces".
$endgroup$
– Christian Blatter
Jan 28 at 9:17
add a comment |
$begingroup$
Compute the area between the surfaces $(x^2+y^2)^2=a^2(x^2-y^2)$ and $x^2+y^2+z^2=a^2.$
My attempt:
I take the first case $zgeq 0$:
$$x=x$$
$$y=y$$
$$z=sqrt{a^2-(x^2+y^2)}$$
then
$$T_x=(1,0,frac {-x}{sqrt{a^2-(x^2+y^2)}})$$
$$T_x=(0,1,frac {-y}{sqrt{a^2-(x^2+y^2)}})$$
Then the area of the surface is:
$$A(S)=int_S dsigma=intint_D frac {a}{sqrt{a^2-(x^2+y^2)}}dxdy$$
By parameterization:
$$x=rcost$$
$$y=rsint$$
then $r=asqrt{cos(2t)}implies tin [0, frac {pi}4]$ and $rin[0,asqrt{cos(2t)}]$
$A(S)=(a^2-sqrt{2})$. But that's a wrong answer, so what am I doing wrong?
calculus integration area
edited Jan 28 at 7:14
BadAtGeometry
298215
asked Jan 28 at 7:08
C. CristiC. Cristi
1,639218
$endgroup$
Compute the area between the surfaces $(x^2+y^2)^2=a^2(x^2-y^2)$ and $x^2+y^2+z^2=a^2.$
My attempt:
I take the first case $zgeq 0$:
$$x=x$$
$$y=y$$
$$z=sqrt{a^2-(x^2+y^2)}$$
then
$$T_x=(1,0,frac {-x}{sqrt{a^2-(x^2+y^2)}})$$
$$T_x=(0,1,frac {-y}{sqrt{a^2-(x^2+y^2)}})$$
Then the area of the surface is:
$$A(S)=int_S dsigma=intint_D frac {a}{sqrt{a^2-(x^2+y^2)}}dxdy$$
By parameterization:
$$x=rcost$$
$$y=rsint$$
then $r=asqrt{cos(2t)}implies tin [0, frac {pi}4]$ and $rin[0,asqrt{cos(2t)}]$
$A(S)=(a^2-sqrt{2})$. But that's a wrong answer, so what am I doing wrong?
calculus integration area
calculus integration area
edited Jan 28 at 7:14
BadAtGeometry
298215
asked Jan 28 at 7:08
C. CristiC. Cristi
1,639218
edited Jan 28 at 7:14
BadAtGeometry
298215
asked Jan 28 at 7:08
C. CristiC. Cristi
1,639218
edited Jan 28 at 7:14
BadAtGeometry
298215
edited Jan 28 at 7:14
BadAtGeometry
298215
edited Jan 28 at 7:14
BadAtGeometry
298215
298215
asked Jan 28 at 7:08
C. CristiC. Cristi
1,639218
asked Jan 28 at 7:08
C. CristiC. Cristi
1,639218
asked Jan 28 at 7:08
C. CristiC. Cristi
1,639218
1,639218
1
$begingroup$
It is unclear what you mean by the "area between the surfaces".
$endgroup$
– Christian Blatter
Jan 28 at 9:17
add a comment |
1
$begingroup$
It is unclear what you mean by the "area between the surfaces".
$endgroup$
– Christian Blatter
Jan 28 at 9:17
1
1
$begingroup$
It is unclear what you mean by the "area between the surfaces".
$endgroup$
– Christian Blatter
Jan 28 at 9:17
$begingroup$
It is unclear what you mean by the "area between the surfaces".
$endgroup$
– Christian Blatter
Jan 28 at 9:17
add a comment |
1 Answer
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oldest
votes
$begingroup$
Using parametric form $x = rcost$ and $y = rsint$ the equation of the curve is $r^2 = a^2cos2t$ and the equation of the sphere is $x = rcost$, $y = rsint$, $z = sqrt{a^2 - r^2}$
The integral of surface area is given by
$$intint_Rsqrt{EG - F^2}drdt$$
$$E = (frac{partial x}{partial r})^2 +(frac{partial y}{partial r})^2 + (frac{partial z}{partial r})^2 = frac{a^2}{a^2 - r^2}$$
$$G = (frac{partial x}{partial t})^2 + (frac{partial y}{partial t})^2 + (frac{partial z}{partial t})^2 = r^2$$
$$F = (frac{partial x}{partial r})(frac{partial x}{partial t}) + (frac{partial y}{partial r})(frac{partial y}{partial t}) + (frac{partial z}{partial r})(frac{partial z}{partial t}) = 0$$
Therefore the integral for the surface area for $y > 0$ at the upper sphere is
$$int_0^{frac{pi}{4}}int_0^{asqrt{cos2t}}sqrt{frac{a^2r^2}{a^2 - r^2}}drdt$$
$$= a^2(frac{pi}{4} + 1 - sqrt{2})$$
Since we have $y < 0$ and the lower sphere plus there are two branches, the total surface area should be 8 times the above. Hence the total surface area is
$$a^2(2pi + 8 - 8sqrt{2})$$
answered Feb 2 at 21:45
KY TangKY Tang
47936
$endgroup$
add a comment |
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$begingroup$
Using parametric form $x = rcost$ and $y = rsint$ the equation of the curve is $r^2 = a^2cos2t$ and the equation of the sphere is $x = rcost$, $y = rsint$, $z = sqrt{a^2 - r^2}$
The integral of surface area is given by
$$intint_Rsqrt{EG - F^2}drdt$$
$$E = (frac{partial x}{partial r})^2 +(frac{partial y}{partial r})^2 + (frac{partial z}{partial r})^2 = frac{a^2}{a^2 - r^2}$$
$$G = (frac{partial x}{partial t})^2 + (frac{partial y}{partial t})^2 + (frac{partial z}{partial t})^2 = r^2$$
$$F = (frac{partial x}{partial r})(frac{partial x}{partial t}) + (frac{partial y}{partial r})(frac{partial y}{partial t}) + (frac{partial z}{partial r})(frac{partial z}{partial t}) = 0$$
Therefore the integral for the surface area for $y > 0$ at the upper sphere is
$$int_0^{frac{pi}{4}}int_0^{asqrt{cos2t}}sqrt{frac{a^2r^2}{a^2 - r^2}}drdt$$
$$= a^2(frac{pi}{4} + 1 - sqrt{2})$$
Since we have $y < 0$ and the lower sphere plus there are two branches, the total surface area should be 8 times the above. Hence the total surface area is
$$a^2(2pi + 8 - 8sqrt{2})$$
answered Feb 2 at 21:45
KY TangKY Tang
47936
$endgroup$
add a comment |
$begingroup$
Using parametric form $x = rcost$ and $y = rsint$ the equation of the curve is $r^2 = a^2cos2t$ and the equation of the sphere is $x = rcost$, $y = rsint$, $z = sqrt{a^2 - r^2}$
The integral of surface area is given by
$$intint_Rsqrt{EG - F^2}drdt$$
$$E = (frac{partial x}{partial r})^2 +(frac{partial y}{partial r})^2 + (frac{partial z}{partial r})^2 = frac{a^2}{a^2 - r^2}$$
$$G = (frac{partial x}{partial t})^2 + (frac{partial y}{partial t})^2 + (frac{partial z}{partial t})^2 = r^2$$
$$F = (frac{partial x}{partial r})(frac{partial x}{partial t}) + (frac{partial y}{partial r})(frac{partial y}{partial t}) + (frac{partial z}{partial r})(frac{partial z}{partial t}) = 0$$
Therefore the integral for the surface area for $y > 0$ at the upper sphere is
$$int_0^{frac{pi}{4}}int_0^{asqrt{cos2t}}sqrt{frac{a^2r^2}{a^2 - r^2}}drdt$$
$$= a^2(frac{pi}{4} + 1 - sqrt{2})$$
Since we have $y < 0$ and the lower sphere plus there are two branches, the total surface area should be 8 times the above. Hence the total surface area is
$$a^2(2pi + 8 - 8sqrt{2})$$
answered Feb 2 at 21:45
KY TangKY Tang
47936
$endgroup$
add a comment |
$begingroup$
Using parametric form $x = rcost$ and $y = rsint$ the equation of the curve is $r^2 = a^2cos2t$ and the equation of the sphere is $x = rcost$, $y = rsint$, $z = sqrt{a^2 - r^2}$
The integral of surface area is given by
$$intint_Rsqrt{EG - F^2}drdt$$
$$E = (frac{partial x}{partial r})^2 +(frac{partial y}{partial r})^2 + (frac{partial z}{partial r})^2 = frac{a^2}{a^2 - r^2}$$
$$G = (frac{partial x}{partial t})^2 + (frac{partial y}{partial t})^2 + (frac{partial z}{partial t})^2 = r^2$$
$$F = (frac{partial x}{partial r})(frac{partial x}{partial t}) + (frac{partial y}{partial r})(frac{partial y}{partial t}) + (frac{partial z}{partial r})(frac{partial z}{partial t}) = 0$$
Therefore the integral for the surface area for $y > 0$ at the upper sphere is
$$int_0^{frac{pi}{4}}int_0^{asqrt{cos2t}}sqrt{frac{a^2r^2}{a^2 - r^2}}drdt$$
$$= a^2(frac{pi}{4} + 1 - sqrt{2})$$
Since we have $y < 0$ and the lower sphere plus there are two branches, the total surface area should be 8 times the above. Hence the total surface area is
$$a^2(2pi + 8 - 8sqrt{2})$$
answered Feb 2 at 21:45
KY TangKY Tang
47936
$endgroup$
Using parametric form $x = rcost$ and $y = rsint$ the equation of the curve is $r^2 = a^2cos2t$ and the equation of the sphere is $x = rcost$, $y = rsint$, $z = sqrt{a^2 - r^2}$
The integral of surface area is given by
$$intint_Rsqrt{EG - F^2}drdt$$
$$E = (frac{partial x}{partial r})^2 +(frac{partial y}{partial r})^2 + (frac{partial z}{partial r})^2 = frac{a^2}{a^2 - r^2}$$
$$G = (frac{partial x}{partial t})^2 + (frac{partial y}{partial t})^2 + (frac{partial z}{partial t})^2 = r^2$$
$$F = (frac{partial x}{partial r})(frac{partial x}{partial t}) + (frac{partial y}{partial r})(frac{partial y}{partial t}) + (frac{partial z}{partial r})(frac{partial z}{partial t}) = 0$$
Therefore the integral for the surface area for $y > 0$ at the upper sphere is
$$int_0^{frac{pi}{4}}int_0^{asqrt{cos2t}}sqrt{frac{a^2r^2}{a^2 - r^2}}drdt$$
$$= a^2(frac{pi}{4} + 1 - sqrt{2})$$
Since we have $y < 0$ and the lower sphere plus there are two branches, the total surface area should be 8 times the above. Hence the total surface area is
$$a^2(2pi + 8 - 8sqrt{2})$$
answered Feb 2 at 21:45
KY TangKY Tang
47936
answered Feb 2 at 21:45
KY TangKY Tang
47936
answered Feb 2 at 21:45
KY TangKY Tang
47936
answered Feb 2 at 21:45
KY TangKY Tang
47936
47936
add a comment |
add a comment |
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$begingroup$
It is unclear what you mean by the "area between the surfaces".
$endgroup$
– Christian Blatter
Jan 28 at 9:17