Area of triangle using double integrals
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I have one (rather simple) problem, but I'm stuck and can't figure out what I'm constantly doing wrong. I need to calculate area of triangle with points at $(0,0)$, $(t,0)$, $(t,frac{t}{2})$. In other words triangle under function $y=frac{x}{2}$, for $xin [0,t]$ I thought it is calculated with
$$ int_0^t int_0^frac{t}{2} dudv$$
But it turns out that this equals to $frac{t^2}{2}$, when obviously this area is $frac{ttimesfrac{t}{2}}{2} = frac{t^2}{4}$.
What am I doing wrong here?
I need to calculate it this way, not with single integral, or geometrically.
integration multivariable-calculus area
edited Dec 4 '18 at 16:25
José Carlos Santos
153k22123225
asked Dec 4 '18 at 16:16
Vasilije DavidovicVasilije Davidovic
132
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add a comment |
$begingroup$
I have one (rather simple) problem, but I'm stuck and can't figure out what I'm constantly doing wrong. I need to calculate area of triangle with points at $(0,0)$, $(t,0)$, $(t,frac{t}{2})$. In other words triangle under function $y=frac{x}{2}$, for $xin [0,t]$ I thought it is calculated with
$$ int_0^t int_0^frac{t}{2} dudv$$
But it turns out that this equals to $frac{t^2}{2}$, when obviously this area is $frac{ttimesfrac{t}{2}}{2} = frac{t^2}{4}$.
What am I doing wrong here?
I need to calculate it this way, not with single integral, or geometrically.
integration multivariable-calculus area
edited Dec 4 '18 at 16:25
José Carlos Santos
153k22123225
asked Dec 4 '18 at 16:16
Vasilije DavidovicVasilije Davidovic
132
$endgroup$
add a comment |
$begingroup$
I have one (rather simple) problem, but I'm stuck and can't figure out what I'm constantly doing wrong. I need to calculate area of triangle with points at $(0,0)$, $(t,0)$, $(t,frac{t}{2})$. In other words triangle under function $y=frac{x}{2}$, for $xin [0,t]$ I thought it is calculated with
$$ int_0^t int_0^frac{t}{2} dudv$$
But it turns out that this equals to $frac{t^2}{2}$, when obviously this area is $frac{ttimesfrac{t}{2}}{2} = frac{t^2}{4}$.
What am I doing wrong here?
I need to calculate it this way, not with single integral, or geometrically.
integration multivariable-calculus area
edited Dec 4 '18 at 16:25
José Carlos Santos
153k22123225
asked Dec 4 '18 at 16:16
Vasilije DavidovicVasilije Davidovic
132
$endgroup$
I have one (rather simple) problem, but I'm stuck and can't figure out what I'm constantly doing wrong. I need to calculate area of triangle with points at $(0,0)$, $(t,0)$, $(t,frac{t}{2})$. In other words triangle under function $y=frac{x}{2}$, for $xin [0,t]$ I thought it is calculated with
$$ int_0^t int_0^frac{t}{2} dudv$$
But it turns out that this equals to $frac{t^2}{2}$, when obviously this area is $frac{ttimesfrac{t}{2}}{2} = frac{t^2}{4}$.
What am I doing wrong here?
I need to calculate it this way, not with single integral, or geometrically.
integration multivariable-calculus area
integration multivariable-calculus area
edited Dec 4 '18 at 16:25
José Carlos Santos
153k22123225
asked Dec 4 '18 at 16:16
Vasilije DavidovicVasilije Davidovic
132
edited Dec 4 '18 at 16:25
José Carlos Santos
153k22123225
asked Dec 4 '18 at 16:16
Vasilije DavidovicVasilije Davidovic
132
edited Dec 4 '18 at 16:25
José Carlos Santos
153k22123225
edited Dec 4 '18 at 16:25
José Carlos Santos
153k22123225
edited Dec 4 '18 at 16:25
José Carlos Santos
153k22123225
153k22123225
asked Dec 4 '18 at 16:16
Vasilije DavidovicVasilije Davidovic
132
asked Dec 4 '18 at 16:16
Vasilije DavidovicVasilije Davidovic
132
asked Dec 4 '18 at 16:16
Vasilije DavidovicVasilije Davidovic
132
132
add a comment |
add a comment |
2 Answers
2
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oldest
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The integral that you actually computed corresponds to the area of a rectangle. You should actually compute$$int_0^tint_0^{frac x2},mathrm dy,mathrm dx.$$
answered Dec 4 '18 at 16:20
José Carlos SantosJosé Carlos Santos
153k22123225
$endgroup$
$begingroup$
Silly me. Names for variables were different in my notes, so I got confused. Thank you very much for your reply.
$endgroup$
– Vasilije Davidovic
Dec 4 '18 at 16:32
add a comment |
$begingroup$
I think you made a mistake, the bounds on the inner integral should not be constant:
$$
int_0^t int_0^{v/2} dudv
= int_0^t frac{v}{2} dv
= left. frac{v^2}{4} right|_0^t = frac{t^2}{4}.
$$
answered Dec 4 '18 at 16:20
gt6989bgt6989b
33.3k22452
$endgroup$
$begingroup$
Thank you very much for your reply. That solved it
$endgroup$
– Vasilije Davidovic
Dec 4 '18 at 16:33
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
$begingroup$
The integral that you actually computed corresponds to the area of a rectangle. You should actually compute$$int_0^tint_0^{frac x2},mathrm dy,mathrm dx.$$
answered Dec 4 '18 at 16:20
José Carlos SantosJosé Carlos Santos
153k22123225
$endgroup$
$begingroup$
Silly me. Names for variables were different in my notes, so I got confused. Thank you very much for your reply.
$endgroup$
– Vasilije Davidovic
Dec 4 '18 at 16:32
add a comment |
$begingroup$
The integral that you actually computed corresponds to the area of a rectangle. You should actually compute$$int_0^tint_0^{frac x2},mathrm dy,mathrm dx.$$
answered Dec 4 '18 at 16:20
José Carlos SantosJosé Carlos Santos
153k22123225
$endgroup$
$begingroup$
Silly me. Names for variables were different in my notes, so I got confused. Thank you very much for your reply.
$endgroup$
– Vasilije Davidovic
Dec 4 '18 at 16:32
add a comment |
$begingroup$
The integral that you actually computed corresponds to the area of a rectangle. You should actually compute$$int_0^tint_0^{frac x2},mathrm dy,mathrm dx.$$
answered Dec 4 '18 at 16:20
José Carlos SantosJosé Carlos Santos
153k22123225
$endgroup$
The integral that you actually computed corresponds to the area of a rectangle. You should actually compute$$int_0^tint_0^{frac x2},mathrm dy,mathrm dx.$$
answered Dec 4 '18 at 16:20
José Carlos SantosJosé Carlos Santos
153k22123225
answered Dec 4 '18 at 16:20
José Carlos SantosJosé Carlos Santos
153k22123225
answered Dec 4 '18 at 16:20
José Carlos SantosJosé Carlos Santos
153k22123225
answered Dec 4 '18 at 16:20
José Carlos SantosJosé Carlos Santos
153k22123225
153k22123225
$begingroup$
Silly me. Names for variables were different in my notes, so I got confused. Thank you very much for your reply.
$endgroup$
– Vasilije Davidovic
Dec 4 '18 at 16:32
add a comment |
$begingroup$
Silly me. Names for variables were different in my notes, so I got confused. Thank you very much for your reply.
$endgroup$
– Vasilije Davidovic
Dec 4 '18 at 16:32
$begingroup$
Silly me. Names for variables were different in my notes, so I got confused. Thank you very much for your reply.
$endgroup$
– Vasilije Davidovic
Dec 4 '18 at 16:32
$begingroup$
Silly me. Names for variables were different in my notes, so I got confused. Thank you very much for your reply.
$endgroup$
– Vasilije Davidovic
Dec 4 '18 at 16:32
add a comment |
$begingroup$
I think you made a mistake, the bounds on the inner integral should not be constant:
$$
int_0^t int_0^{v/2} dudv
= int_0^t frac{v}{2} dv
= left. frac{v^2}{4} right|_0^t = frac{t^2}{4}.
$$
answered Dec 4 '18 at 16:20
gt6989bgt6989b
33.3k22452
$endgroup$
$begingroup$
Thank you very much for your reply. That solved it
$endgroup$
– Vasilije Davidovic
Dec 4 '18 at 16:33
add a comment |
$begingroup$
I think you made a mistake, the bounds on the inner integral should not be constant:
$$
int_0^t int_0^{v/2} dudv
= int_0^t frac{v}{2} dv
= left. frac{v^2}{4} right|_0^t = frac{t^2}{4}.
$$
answered Dec 4 '18 at 16:20
gt6989bgt6989b
33.3k22452
$endgroup$
$begingroup$
Thank you very much for your reply. That solved it
$endgroup$
– Vasilije Davidovic
Dec 4 '18 at 16:33
add a comment |
$begingroup$
I think you made a mistake, the bounds on the inner integral should not be constant:
$$
int_0^t int_0^{v/2} dudv
= int_0^t frac{v}{2} dv
= left. frac{v^2}{4} right|_0^t = frac{t^2}{4}.
$$
answered Dec 4 '18 at 16:20
gt6989bgt6989b
33.3k22452
$endgroup$
I think you made a mistake, the bounds on the inner integral should not be constant:
$$
int_0^t int_0^{v/2} dudv
= int_0^t frac{v}{2} dv
= left. frac{v^2}{4} right|_0^t = frac{t^2}{4}.
$$
answered Dec 4 '18 at 16:20
gt6989bgt6989b
33.3k22452
answered Dec 4 '18 at 16:20
gt6989bgt6989b
33.3k22452
answered Dec 4 '18 at 16:20
gt6989bgt6989b
33.3k22452
answered Dec 4 '18 at 16:20
gt6989bgt6989b
33.3k22452
33.3k22452
$begingroup$
Thank you very much for your reply. That solved it
$endgroup$
– Vasilije Davidovic
Dec 4 '18 at 16:33
add a comment |
$begingroup$
Thank you very much for your reply. That solved it
$endgroup$
– Vasilije Davidovic
Dec 4 '18 at 16:33
$begingroup$
Thank you very much for your reply. That solved it
$endgroup$
– Vasilije Davidovic
Dec 4 '18 at 16:33
$begingroup$
Thank you very much for your reply. That solved it
$endgroup$
– Vasilije Davidovic
Dec 4 '18 at 16:33
add a comment |
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