Find the area of the surface formed by revolving the given curve about $(i)x$-axis and $(i)y$-axis
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Q:Find the area of the surface formed by revolving the given curve about $(i)x-axis$ and $(i)y-axis$
$$x=acostheta ,y=bsintheta,0lethetale2pi$$
About $x-$axis is, $S=2piint_0^{2pi}bsintheta sqrt{a^2(sintheta)^2+b^2(costheta)^2} dtheta$
About $y-$axis is, $S=2piint_0^{2pi}acostheta sqrt{a^2(sintheta)^2+b^2(costheta)^2} dtheta$
from now i get stuck.I can't not figure out the integral part.Any hints or solution will be appreciated.
Thanks in advance.
calculus parametric solid-of-revolution
edited Nov 23 at 12:51
Wesley Strik
1,510422
asked Nov 23 at 11:44
raihan hossain
818
add a comment |
up vote
1
down vote
favorite
Q:Find the area of the surface formed by revolving the given curve about $(i)x-axis$ and $(i)y-axis$
$$x=acostheta ,y=bsintheta,0lethetale2pi$$
About $x-$axis is, $S=2piint_0^{2pi}bsintheta sqrt{a^2(sintheta)^2+b^2(costheta)^2} dtheta$
About $y-$axis is, $S=2piint_0^{2pi}acostheta sqrt{a^2(sintheta)^2+b^2(costheta)^2} dtheta$
from now i get stuck.I can't not figure out the integral part.Any hints or solution will be appreciated.
Thanks in advance.
calculus parametric solid-of-revolution
edited Nov 23 at 12:51
Wesley Strik
1,510422
asked Nov 23 at 11:44
raihan hossain
818
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Q:Find the area of the surface formed by revolving the given curve about $(i)x-axis$ and $(i)y-axis$
$$x=acostheta ,y=bsintheta,0lethetale2pi$$
About $x-$axis is, $S=2piint_0^{2pi}bsintheta sqrt{a^2(sintheta)^2+b^2(costheta)^2} dtheta$
About $y-$axis is, $S=2piint_0^{2pi}acostheta sqrt{a^2(sintheta)^2+b^2(costheta)^2} dtheta$
from now i get stuck.I can't not figure out the integral part.Any hints or solution will be appreciated.
Thanks in advance.
calculus parametric solid-of-revolution
edited Nov 23 at 12:51
Wesley Strik
1,510422
asked Nov 23 at 11:44
raihan hossain
818
Q:Find the area of the surface formed by revolving the given curve about $(i)x-axis$ and $(i)y-axis$
$$x=acostheta ,y=bsintheta,0lethetale2pi$$
About $x-$axis is, $S=2piint_0^{2pi}bsintheta sqrt{a^2(sintheta)^2+b^2(costheta)^2} dtheta$
About $y-$axis is, $S=2piint_0^{2pi}acostheta sqrt{a^2(sintheta)^2+b^2(costheta)^2} dtheta$
from now i get stuck.I can't not figure out the integral part.Any hints or solution will be appreciated.
Thanks in advance.
calculus parametric solid-of-revolution
calculus parametric solid-of-revolution
edited Nov 23 at 12:51
Wesley Strik
1,510422
asked Nov 23 at 11:44
raihan hossain
818
edited Nov 23 at 12:51
Wesley Strik
1,510422
asked Nov 23 at 11:44
raihan hossain
818
edited Nov 23 at 12:51
Wesley Strik
1,510422
edited Nov 23 at 12:51
Wesley Strik
1,510422
edited Nov 23 at 12:51
Wesley Strik
1,510422
1,510422
asked Nov 23 at 11:44
raihan hossain
818
asked Nov 23 at 11:44
raihan hossain
818
asked Nov 23 at 11:44
raihan hossain
818
818
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
The limits of integration need some correction. While finding the surface area about the $x$-axis, $x$ ranges from $-a$ to $aimpliestheta$ ranges from $pirightarrow 2pi$, not $0rightarrow 2pi$. For the surface area about the $y$-axis, $theta$ ranges from $-pi/2 rightarrow +pi/2$, or from $3pi/2rightarrow 2pi$ and $0rightarrowpi/2$.
For the surface area about $x$-axis, take $t=costheta implies dt=-sintheta dtheta$
$S_x=2piint_pi^{2pi}|bsintheta| sqrt{a^2sin^2theta+b^2cos^2theta} dtheta\ =2pi bint_pi^{2pi}|sintheta| sqrt{a^2(1-cos^2theta)+b^2cos^2theta} dtheta\\ =2pi bint_pi^{2pi}(-sintheta) sqrt{a^2+(b^2-a^2)cos^2theta} dtheta\\ =2pi bint_{-1}^{1}sqrt{a^2+(b^2-a^2)t^2} dt\\ =4pi bint_0^{1}sqrt{a^2+(b^2-a^2)t^2} dt\$
Depending on the sign of $(b^2-a^2)$, this integral can take either of the standard forms $int sqrt{a^2-x^2} dx$ or $int sqrt{a^2+x^2} dx$.
For the surface area about $y$-axis, because we have $costheta dtheta$ outside the square root, take $t=sintheta$, and try to get the argument of the square root in terms of $sintheta$ alone, this time by substituting for $cos^2theta$.
answered Nov 23 at 13:14
Shubham Johri
2,969413
Thanks @Shubham Johri Sir for your nice explanation :) but I have a question:How could I know the sign of $(b^2-a^2)$? Because the question isn't provide anything.Can I show both standard form for the seek of general answer??! Thanks again
– raihan hossain
Nov 23 at 13:34
You may use both the forms to get a general answer that takes care of both the cases $a>b$ and $b>a$.
– Shubham Johri
Nov 23 at 13:50
add a comment |
up vote
2
down vote
Hint: For the first one let $costheta=u$
$$
S=2piint_0^{pi}bsintheta sqrt{a^2(sintheta)^2+b^2(costheta)^2} dtheta
=2piint_{-1}^1 bsqrt{a^2+(b^2-a^2)u^2} du
$$
and then let substitution $sqrt{b^2-a^2}u=atanphi$.
answered Nov 23 at 12:25
Nosrati
26.4k62353
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The limits of integration need some correction. While finding the surface area about the $x$-axis, $x$ ranges from $-a$ to $aimpliestheta$ ranges from $pirightarrow 2pi$, not $0rightarrow 2pi$. For the surface area about the $y$-axis, $theta$ ranges from $-pi/2 rightarrow +pi/2$, or from $3pi/2rightarrow 2pi$ and $0rightarrowpi/2$.
For the surface area about $x$-axis, take $t=costheta implies dt=-sintheta dtheta$
$S_x=2piint_pi^{2pi}|bsintheta| sqrt{a^2sin^2theta+b^2cos^2theta} dtheta\ =2pi bint_pi^{2pi}|sintheta| sqrt{a^2(1-cos^2theta)+b^2cos^2theta} dtheta\\ =2pi bint_pi^{2pi}(-sintheta) sqrt{a^2+(b^2-a^2)cos^2theta} dtheta\\ =2pi bint_{-1}^{1}sqrt{a^2+(b^2-a^2)t^2} dt\\ =4pi bint_0^{1}sqrt{a^2+(b^2-a^2)t^2} dt\$
Depending on the sign of $(b^2-a^2)$, this integral can take either of the standard forms $int sqrt{a^2-x^2} dx$ or $int sqrt{a^2+x^2} dx$.
For the surface area about $y$-axis, because we have $costheta dtheta$ outside the square root, take $t=sintheta$, and try to get the argument of the square root in terms of $sintheta$ alone, this time by substituting for $cos^2theta$.
answered Nov 23 at 13:14
Shubham Johri
2,969413
Thanks @Shubham Johri Sir for your nice explanation :) but I have a question:How could I know the sign of $(b^2-a^2)$? Because the question isn't provide anything.Can I show both standard form for the seek of general answer??! Thanks again
– raihan hossain
Nov 23 at 13:34
You may use both the forms to get a general answer that takes care of both the cases $a>b$ and $b>a$.
– Shubham Johri
Nov 23 at 13:50
add a comment |
up vote
1
down vote
accepted
The limits of integration need some correction. While finding the surface area about the $x$-axis, $x$ ranges from $-a$ to $aimpliestheta$ ranges from $pirightarrow 2pi$, not $0rightarrow 2pi$. For the surface area about the $y$-axis, $theta$ ranges from $-pi/2 rightarrow +pi/2$, or from $3pi/2rightarrow 2pi$ and $0rightarrowpi/2$.
For the surface area about $x$-axis, take $t=costheta implies dt=-sintheta dtheta$
$S_x=2piint_pi^{2pi}|bsintheta| sqrt{a^2sin^2theta+b^2cos^2theta} dtheta\ =2pi bint_pi^{2pi}|sintheta| sqrt{a^2(1-cos^2theta)+b^2cos^2theta} dtheta\\ =2pi bint_pi^{2pi}(-sintheta) sqrt{a^2+(b^2-a^2)cos^2theta} dtheta\\ =2pi bint_{-1}^{1}sqrt{a^2+(b^2-a^2)t^2} dt\\ =4pi bint_0^{1}sqrt{a^2+(b^2-a^2)t^2} dt\$
Depending on the sign of $(b^2-a^2)$, this integral can take either of the standard forms $int sqrt{a^2-x^2} dx$ or $int sqrt{a^2+x^2} dx$.
For the surface area about $y$-axis, because we have $costheta dtheta$ outside the square root, take $t=sintheta$, and try to get the argument of the square root in terms of $sintheta$ alone, this time by substituting for $cos^2theta$.
answered Nov 23 at 13:14
Shubham Johri
2,969413
Thanks @Shubham Johri Sir for your nice explanation :) but I have a question:How could I know the sign of $(b^2-a^2)$? Because the question isn't provide anything.Can I show both standard form for the seek of general answer??! Thanks again
– raihan hossain
Nov 23 at 13:34
You may use both the forms to get a general answer that takes care of both the cases $a>b$ and $b>a$.
– Shubham Johri
Nov 23 at 13:50
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The limits of integration need some correction. While finding the surface area about the $x$-axis, $x$ ranges from $-a$ to $aimpliestheta$ ranges from $pirightarrow 2pi$, not $0rightarrow 2pi$. For the surface area about the $y$-axis, $theta$ ranges from $-pi/2 rightarrow +pi/2$, or from $3pi/2rightarrow 2pi$ and $0rightarrowpi/2$.
For the surface area about $x$-axis, take $t=costheta implies dt=-sintheta dtheta$
$S_x=2piint_pi^{2pi}|bsintheta| sqrt{a^2sin^2theta+b^2cos^2theta} dtheta\ =2pi bint_pi^{2pi}|sintheta| sqrt{a^2(1-cos^2theta)+b^2cos^2theta} dtheta\\ =2pi bint_pi^{2pi}(-sintheta) sqrt{a^2+(b^2-a^2)cos^2theta} dtheta\\ =2pi bint_{-1}^{1}sqrt{a^2+(b^2-a^2)t^2} dt\\ =4pi bint_0^{1}sqrt{a^2+(b^2-a^2)t^2} dt\$
Depending on the sign of $(b^2-a^2)$, this integral can take either of the standard forms $int sqrt{a^2-x^2} dx$ or $int sqrt{a^2+x^2} dx$.
For the surface area about $y$-axis, because we have $costheta dtheta$ outside the square root, take $t=sintheta$, and try to get the argument of the square root in terms of $sintheta$ alone, this time by substituting for $cos^2theta$.
answered Nov 23 at 13:14
Shubham Johri
2,969413
The limits of integration need some correction. While finding the surface area about the $x$-axis, $x$ ranges from $-a$ to $aimpliestheta$ ranges from $pirightarrow 2pi$, not $0rightarrow 2pi$. For the surface area about the $y$-axis, $theta$ ranges from $-pi/2 rightarrow +pi/2$, or from $3pi/2rightarrow 2pi$ and $0rightarrowpi/2$.
For the surface area about $x$-axis, take $t=costheta implies dt=-sintheta dtheta$
$S_x=2piint_pi^{2pi}|bsintheta| sqrt{a^2sin^2theta+b^2cos^2theta} dtheta\ =2pi bint_pi^{2pi}|sintheta| sqrt{a^2(1-cos^2theta)+b^2cos^2theta} dtheta\\ =2pi bint_pi^{2pi}(-sintheta) sqrt{a^2+(b^2-a^2)cos^2theta} dtheta\\ =2pi bint_{-1}^{1}sqrt{a^2+(b^2-a^2)t^2} dt\\ =4pi bint_0^{1}sqrt{a^2+(b^2-a^2)t^2} dt\$
Depending on the sign of $(b^2-a^2)$, this integral can take either of the standard forms $int sqrt{a^2-x^2} dx$ or $int sqrt{a^2+x^2} dx$.
For the surface area about $y$-axis, because we have $costheta dtheta$ outside the square root, take $t=sintheta$, and try to get the argument of the square root in terms of $sintheta$ alone, this time by substituting for $cos^2theta$.
answered Nov 23 at 13:14
Shubham Johri
2,969413
edited Nov 23 at 13:55
answered Nov 23 at 13:14
Shubham Johri
2,969413
answered Nov 23 at 13:14
Shubham Johri
2,969413
answered Nov 23 at 13:14
Shubham Johri
2,969413
2,969413
Thanks @Shubham Johri Sir for your nice explanation :) but I have a question:How could I know the sign of $(b^2-a^2)$? Because the question isn't provide anything.Can I show both standard form for the seek of general answer??! Thanks again
– raihan hossain
Nov 23 at 13:34
You may use both the forms to get a general answer that takes care of both the cases $a>b$ and $b>a$.
– Shubham Johri
Nov 23 at 13:50
add a comment |
Thanks @Shubham Johri Sir for your nice explanation :) but I have a question:How could I know the sign of $(b^2-a^2)$? Because the question isn't provide anything.Can I show both standard form for the seek of general answer??! Thanks again
– raihan hossain
Nov 23 at 13:34
You may use both the forms to get a general answer that takes care of both the cases $a>b$ and $b>a$.
– Shubham Johri
Nov 23 at 13:50
Thanks @Shubham Johri Sir for your nice explanation :) but I have a question:How could I know the sign of $(b^2-a^2)$? Because the question isn't provide anything.Can I show both standard form for the seek of general answer??! Thanks again
– raihan hossain
Nov 23 at 13:34
Thanks @Shubham Johri Sir for your nice explanation :) but I have a question:How could I know the sign of $(b^2-a^2)$? Because the question isn't provide anything.Can I show both standard form for the seek of general answer??! Thanks again
– raihan hossain
Nov 23 at 13:34
You may use both the forms to get a general answer that takes care of both the cases $a>b$ and $b>a$.
– Shubham Johri
Nov 23 at 13:50
You may use both the forms to get a general answer that takes care of both the cases $a>b$ and $b>a$.
– Shubham Johri
Nov 23 at 13:50
add a comment |
up vote
2
down vote
Hint: For the first one let $costheta=u$
$$
S=2piint_0^{pi}bsintheta sqrt{a^2(sintheta)^2+b^2(costheta)^2} dtheta
=2piint_{-1}^1 bsqrt{a^2+(b^2-a^2)u^2} du
$$
and then let substitution $sqrt{b^2-a^2}u=atanphi$.
answered Nov 23 at 12:25
Nosrati
26.4k62353
add a comment |
up vote
2
down vote
Hint: For the first one let $costheta=u$
$$
S=2piint_0^{pi}bsintheta sqrt{a^2(sintheta)^2+b^2(costheta)^2} dtheta
=2piint_{-1}^1 bsqrt{a^2+(b^2-a^2)u^2} du
$$
and then let substitution $sqrt{b^2-a^2}u=atanphi$.
answered Nov 23 at 12:25
Nosrati
26.4k62353
add a comment |
up vote
2
down vote
up vote
2
down vote
Hint: For the first one let $costheta=u$
$$
S=2piint_0^{pi}bsintheta sqrt{a^2(sintheta)^2+b^2(costheta)^2} dtheta
=2piint_{-1}^1 bsqrt{a^2+(b^2-a^2)u^2} du
$$
and then let substitution $sqrt{b^2-a^2}u=atanphi$.
answered Nov 23 at 12:25
Nosrati
26.4k62353
Hint: For the first one let $costheta=u$
$$
S=2piint_0^{pi}bsintheta sqrt{a^2(sintheta)^2+b^2(costheta)^2} dtheta
=2piint_{-1}^1 bsqrt{a^2+(b^2-a^2)u^2} du
$$
and then let substitution $sqrt{b^2-a^2}u=atanphi$.
answered Nov 23 at 12:25
Nosrati
26.4k62353
edited Nov 23 at 12:36
answered Nov 23 at 12:25
Nosrati
26.4k62353
answered Nov 23 at 12:25
Nosrati
26.4k62353
answered Nov 23 at 12:25
Nosrati
26.4k62353
26.4k62353
add a comment |
add a comment |
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