Mixing
Integration with substitution to cylindrical coordinate
$begingroup$
Solve the integral
$$
\A:= int_{{x^2over a^2}+{y^2over b^2}+{z^2over c^2}leq 1}{x^2over a^2}+{y^2over b^2}+{z^2over c^2}dxdydz
$$
In my solution I have substituted to cylindrical coordinates ($x=roperatorname{cos} t,y=roperatorname{sin} t,z=h$) with the ranges $-1leq hleq 1$ and $0leq tleq 2pi$. And when I get fixed $h$ and $t$, then $r$ "moves" from $0$ until it meets a new elipse which determinated by $a'={{hover c}cdot a}, b'={{hover c}cdot b}$. So I need to find the meeting point of $r$ with the new elipse $1={x^2over a'^2}+{y^2over b'^2}=({r cos tover ah})^2+({rsin tover bh})^2$. Hence, if I sign $lambda:=+{abhover sqrt{a^2sin^2 t+b^2cos^2 t}}$, then the required range of $r$ is from $0$ to $lambda$. Hence, ($J=r$),
$$
\ A=int_{-1}^1int_0^{2pi}int_0^lambda r({r^2cos^2 tover a^2} +{r^2sin^2 tover b^2}+{h^2over c^2})drdtdh
\ =...=int_{-1}^1int_0^{2pi}(lambda^4({cos^2 tover 4a^2}+{sin^2 tover 4 b^2})+lambda^2{h^2over c^2})dtdh
\=...=int_{-1}^1int_0^{2pi}({a^6b^6c^2h^8over 4(a^2sin^2 t+b^2 cos^2 t)^3}+{a^4b^4h^6over c^2(a^2sin^2 t+b^2cos^2 t)^2})dtdh
$$
Is this integral doable or I didn't do this exercise right? Thianks
integration multivariable-calculus cylindrical-coordinates
asked Jan 7 at 16:59
J. DoeJ. Doe
14110
$endgroup$
add a comment |
$begingroup$
Solve the integral
$$
\A:= int_{{x^2over a^2}+{y^2over b^2}+{z^2over c^2}leq 1}{x^2over a^2}+{y^2over b^2}+{z^2over c^2}dxdydz
$$
In my solution I have substituted to cylindrical coordinates ($x=roperatorname{cos} t,y=roperatorname{sin} t,z=h$) with the ranges $-1leq hleq 1$ and $0leq tleq 2pi$. And when I get fixed $h$ and $t$, then $r$ "moves" from $0$ until it meets a new elipse which determinated by $a'={{hover c}cdot a}, b'={{hover c}cdot b}$. So I need to find the meeting point of $r$ with the new elipse $1={x^2over a'^2}+{y^2over b'^2}=({r cos tover ah})^2+({rsin tover bh})^2$. Hence, if I sign $lambda:=+{abhover sqrt{a^2sin^2 t+b^2cos^2 t}}$, then the required range of $r$ is from $0$ to $lambda$. Hence, ($J=r$),
$$
\ A=int_{-1}^1int_0^{2pi}int_0^lambda r({r^2cos^2 tover a^2} +{r^2sin^2 tover b^2}+{h^2over c^2})drdtdh
\ =...=int_{-1}^1int_0^{2pi}(lambda^4({cos^2 tover 4a^2}+{sin^2 tover 4 b^2})+lambda^2{h^2over c^2})dtdh
\=...=int_{-1}^1int_0^{2pi}({a^6b^6c^2h^8over 4(a^2sin^2 t+b^2 cos^2 t)^3}+{a^4b^4h^6over c^2(a^2sin^2 t+b^2cos^2 t)^2})dtdh
$$
Is this integral doable or I didn't do this exercise right? Thianks
integration multivariable-calculus cylindrical-coordinates
asked Jan 7 at 16:59
J. DoeJ. Doe
14110
$endgroup$
2
$begingroup$
Where is the variable $h$ coming from? Also if you put a backslash before $sin$ and $cos$, they turn into $sin{t}$ and $cos{t}$.
$endgroup$
– Calvin Godfrey
Jan 7 at 17:03
$begingroup$
@CalvinGodfrey $h$ equals $z$, I've fixed it.
$endgroup$
– J. Doe
Jan 7 at 17:04
2
$begingroup$
Change first to $u=x/a$, $v=y/b$, $w=z/c$ and then to spherical coordinates.
$endgroup$
– A.Γ.
Jan 7 at 17:06
add a comment |
$begingroup$
Solve the integral
$$
\A:= int_{{x^2over a^2}+{y^2over b^2}+{z^2over c^2}leq 1}{x^2over a^2}+{y^2over b^2}+{z^2over c^2}dxdydz
$$
In my solution I have substituted to cylindrical coordinates ($x=roperatorname{cos} t,y=roperatorname{sin} t,z=h$) with the ranges $-1leq hleq 1$ and $0leq tleq 2pi$. And when I get fixed $h$ and $t$, then $r$ "moves" from $0$ until it meets a new elipse which determinated by $a'={{hover c}cdot a}, b'={{hover c}cdot b}$. So I need to find the meeting point of $r$ with the new elipse $1={x^2over a'^2}+{y^2over b'^2}=({r cos tover ah})^2+({rsin tover bh})^2$. Hence, if I sign $lambda:=+{abhover sqrt{a^2sin^2 t+b^2cos^2 t}}$, then the required range of $r$ is from $0$ to $lambda$. Hence, ($J=r$),
$$
\ A=int_{-1}^1int_0^{2pi}int_0^lambda r({r^2cos^2 tover a^2} +{r^2sin^2 tover b^2}+{h^2over c^2})drdtdh
\ =...=int_{-1}^1int_0^{2pi}(lambda^4({cos^2 tover 4a^2}+{sin^2 tover 4 b^2})+lambda^2{h^2over c^2})dtdh
\=...=int_{-1}^1int_0^{2pi}({a^6b^6c^2h^8over 4(a^2sin^2 t+b^2 cos^2 t)^3}+{a^4b^4h^6over c^2(a^2sin^2 t+b^2cos^2 t)^2})dtdh
$$
Is this integral doable or I didn't do this exercise right? Thianks
integration multivariable-calculus cylindrical-coordinates
asked Jan 7 at 16:59
J. DoeJ. Doe
14110
$endgroup$
Solve the integral
$$
\A:= int_{{x^2over a^2}+{y^2over b^2}+{z^2over c^2}leq 1}{x^2over a^2}+{y^2over b^2}+{z^2over c^2}dxdydz
$$
In my solution I have substituted to cylindrical coordinates ($x=roperatorname{cos} t,y=roperatorname{sin} t,z=h$) with the ranges $-1leq hleq 1$ and $0leq tleq 2pi$. And when I get fixed $h$ and $t$, then $r$ "moves" from $0$ until it meets a new elipse which determinated by $a'={{hover c}cdot a}, b'={{hover c}cdot b}$. So I need to find the meeting point of $r$ with the new elipse $1={x^2over a'^2}+{y^2over b'^2}=({r cos tover ah})^2+({rsin tover bh})^2$. Hence, if I sign $lambda:=+{abhover sqrt{a^2sin^2 t+b^2cos^2 t}}$, then the required range of $r$ is from $0$ to $lambda$. Hence, ($J=r$),
$$
\ A=int_{-1}^1int_0^{2pi}int_0^lambda r({r^2cos^2 tover a^2} +{r^2sin^2 tover b^2}+{h^2over c^2})drdtdh
\ =...=int_{-1}^1int_0^{2pi}(lambda^4({cos^2 tover 4a^2}+{sin^2 tover 4 b^2})+lambda^2{h^2over c^2})dtdh
\=...=int_{-1}^1int_0^{2pi}({a^6b^6c^2h^8over 4(a^2sin^2 t+b^2 cos^2 t)^3}+{a^4b^4h^6over c^2(a^2sin^2 t+b^2cos^2 t)^2})dtdh
$$
Is this integral doable or I didn't do this exercise right? Thianks
integration multivariable-calculus cylindrical-coordinates
integration multivariable-calculus cylindrical-coordinates
asked Jan 7 at 16:59
J. DoeJ. Doe
14110
asked Jan 7 at 16:59
J. DoeJ. Doe
14110
edited Jan 7 at 17:04
J. Doe
asked Jan 7 at 16:59
J. DoeJ. Doe
14110
asked Jan 7 at 16:59
J. DoeJ. Doe
14110
asked Jan 7 at 16:59
J. DoeJ. Doe
14110
14110
2
$begingroup$
Where is the variable $h$ coming from? Also if you put a backslash before $sin$ and $cos$, they turn into $sin{t}$ and $cos{t}$.
$endgroup$
– Calvin Godfrey
Jan 7 at 17:03
$begingroup$
@CalvinGodfrey $h$ equals $z$, I've fixed it.
$endgroup$
– J. Doe
Jan 7 at 17:04
2
$begingroup$
Change first to $u=x/a$, $v=y/b$, $w=z/c$ and then to spherical coordinates.
$endgroup$
– A.Γ.
Jan 7 at 17:06
add a comment |
2
$begingroup$
Where is the variable $h$ coming from? Also if you put a backslash before $sin$ and $cos$, they turn into $sin{t}$ and $cos{t}$.
$endgroup$
– Calvin Godfrey
Jan 7 at 17:03
$begingroup$
@CalvinGodfrey $h$ equals $z$, I've fixed it.
$endgroup$
– J. Doe
Jan 7 at 17:04
2
$begingroup$
Change first to $u=x/a$, $v=y/b$, $w=z/c$ and then to spherical coordinates.
$endgroup$
– A.Γ.
Jan 7 at 17:06
2
2
$begingroup$
Where is the variable $h$ coming from? Also if you put a backslash before $sin$ and $cos$, they turn into $sin{t}$ and $cos{t}$.
$endgroup$
– Calvin Godfrey
Jan 7 at 17:03
$begingroup$
Where is the variable $h$ coming from? Also if you put a backslash before $sin$ and $cos$, they turn into $sin{t}$ and $cos{t}$.
$endgroup$
– Calvin Godfrey
Jan 7 at 17:03
$begingroup$
@CalvinGodfrey $h$ equals $z$, I've fixed it.
$endgroup$
– J. Doe
Jan 7 at 17:04
$begingroup$
@CalvinGodfrey $h$ equals $z$, I've fixed it.
$endgroup$
– J. Doe
Jan 7 at 17:04
2
2
$begingroup$
Change first to $u=x/a$, $v=y/b$, $w=z/c$ and then to spherical coordinates.
$endgroup$
– A.Γ.
Jan 7 at 17:06
$begingroup$
Change first to $u=x/a$, $v=y/b$, $w=z/c$ and then to spherical coordinates.
$endgroup$
– A.Γ.
Jan 7 at 17:06
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint for a simpler way. Let $X=x/a$, $Y=y/b$ and $Z=z/c$, then
$$begin{align}int_{{x^2over a^2}+{y^2over b^2}+{z^2over c^2}leq 1}&left({x^2over a^2}+{y^2over b^2}+{z^2over c^2}right)dxdydz\
&=
abcint_{{X^2}+{Y^2}+{Z^2}leq 1}left({X^2}+{Y^2}+{Z^2}right)dXdYdZ\
&=abcint_{r=0}^1r^2 (4pi r^2) dr
end{align}$$
where in the last step we used the spherical coordinates.
answered Jan 7 at 17:07
Robert ZRobert Z
95.8k1065136
$endgroup$
add a comment |
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1 Answer
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active
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1 Answer
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active
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$begingroup$
Hint for a simpler way. Let $X=x/a$, $Y=y/b$ and $Z=z/c$, then
$$begin{align}int_{{x^2over a^2}+{y^2over b^2}+{z^2over c^2}leq 1}&left({x^2over a^2}+{y^2over b^2}+{z^2over c^2}right)dxdydz\
&=
abcint_{{X^2}+{Y^2}+{Z^2}leq 1}left({X^2}+{Y^2}+{Z^2}right)dXdYdZ\
&=abcint_{r=0}^1r^2 (4pi r^2) dr
end{align}$$
where in the last step we used the spherical coordinates.
answered Jan 7 at 17:07
Robert ZRobert Z
95.8k1065136
$endgroup$
add a comment |
$begingroup$
Hint for a simpler way. Let $X=x/a$, $Y=y/b$ and $Z=z/c$, then
$$begin{align}int_{{x^2over a^2}+{y^2over b^2}+{z^2over c^2}leq 1}&left({x^2over a^2}+{y^2over b^2}+{z^2over c^2}right)dxdydz\
&=
abcint_{{X^2}+{Y^2}+{Z^2}leq 1}left({X^2}+{Y^2}+{Z^2}right)dXdYdZ\
&=abcint_{r=0}^1r^2 (4pi r^2) dr
end{align}$$
where in the last step we used the spherical coordinates.
answered Jan 7 at 17:07
Robert ZRobert Z
95.8k1065136
$endgroup$
add a comment |
$begingroup$
Hint for a simpler way. Let $X=x/a$, $Y=y/b$ and $Z=z/c$, then
$$begin{align}int_{{x^2over a^2}+{y^2over b^2}+{z^2over c^2}leq 1}&left({x^2over a^2}+{y^2over b^2}+{z^2over c^2}right)dxdydz\
&=
abcint_{{X^2}+{Y^2}+{Z^2}leq 1}left({X^2}+{Y^2}+{Z^2}right)dXdYdZ\
&=abcint_{r=0}^1r^2 (4pi r^2) dr
end{align}$$
where in the last step we used the spherical coordinates.
answered Jan 7 at 17:07
Robert ZRobert Z
95.8k1065136
$endgroup$
Hint for a simpler way. Let $X=x/a$, $Y=y/b$ and $Z=z/c$, then
$$begin{align}int_{{x^2over a^2}+{y^2over b^2}+{z^2over c^2}leq 1}&left({x^2over a^2}+{y^2over b^2}+{z^2over c^2}right)dxdydz\
&=
abcint_{{X^2}+{Y^2}+{Z^2}leq 1}left({X^2}+{Y^2}+{Z^2}right)dXdYdZ\
&=abcint_{r=0}^1r^2 (4pi r^2) dr
end{align}$$
where in the last step we used the spherical coordinates.
answered Jan 7 at 17:07
Robert ZRobert Z
95.8k1065136
edited Jan 7 at 17:14
answered Jan 7 at 17:07
Robert ZRobert Z
95.8k1065136
answered Jan 7 at 17:07
Robert ZRobert Z
95.8k1065136
answered Jan 7 at 17:07
Robert ZRobert Z
95.8k1065136
95.8k1065136
add a comment |
add a comment |
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2
$begingroup$
Where is the variable $h$ coming from? Also if you put a backslash before $sin$ and $cos$, they turn into $sin{t}$ and $cos{t}$.
$endgroup$
– Calvin Godfrey
Jan 7 at 17:03
$begingroup$
@CalvinGodfrey $h$ equals $z$, I've fixed it.
$endgroup$
– J. Doe
Jan 7 at 17:04
2
$begingroup$
Change first to $u=x/a$, $v=y/b$, $w=z/c$ and then to spherical coordinates.
$endgroup$
– A.Γ.
Jan 7 at 17:06