Calculating limits of integrals of volume
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I have a few problems to do with finding the volume under surfaces. The first ones were fairly simple:
$$h(x,y)=xy^2,,0le xle1,,1le yle2$$
$$therefore V=int_1^2int_0^1int_0^{xy^2}dzdxdy=int_1^2int_0^1xy^2dxdy=frac{1}{2}int_1^2y^2dy=frac{7}{6}$$
However, on later questions the boundaries are given as:
$$h(x,y)=e^{-x-y},,0le x,,y<infty$$
which to me implies:
$$V=int_{-infty}^inftyint_0^inftyint_0^{exp(-x-y)}dzdxdy.$$
However this evaluates to:
$$V=left[e^yright]_{-infty}^infty.$$
which is clearly divergent.
How should I calculate what the limits are for this volume?
calculus integration definite-integrals
edited Nov 13 at 13:36
Ernie060
2,520319
asked Nov 13 at 13:30
Henry Lee
1,626117
add a comment |
up vote
0
down vote
favorite
I have a few problems to do with finding the volume under surfaces. The first ones were fairly simple:
$$h(x,y)=xy^2,,0le xle1,,1le yle2$$
$$therefore V=int_1^2int_0^1int_0^{xy^2}dzdxdy=int_1^2int_0^1xy^2dxdy=frac{1}{2}int_1^2y^2dy=frac{7}{6}$$
However, on later questions the boundaries are given as:
$$h(x,y)=e^{-x-y},,0le x,,y<infty$$
which to me implies:
$$V=int_{-infty}^inftyint_0^inftyint_0^{exp(-x-y)}dzdxdy.$$
However this evaluates to:
$$V=left[e^yright]_{-infty}^infty.$$
which is clearly divergent.
How should I calculate what the limits are for this volume?
calculus integration definite-integrals
edited Nov 13 at 13:36
Ernie060
2,520319
asked Nov 13 at 13:30
Henry Lee
1,626117
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have a few problems to do with finding the volume under surfaces. The first ones were fairly simple:
$$h(x,y)=xy^2,,0le xle1,,1le yle2$$
$$therefore V=int_1^2int_0^1int_0^{xy^2}dzdxdy=int_1^2int_0^1xy^2dxdy=frac{1}{2}int_1^2y^2dy=frac{7}{6}$$
However, on later questions the boundaries are given as:
$$h(x,y)=e^{-x-y},,0le x,,y<infty$$
which to me implies:
$$V=int_{-infty}^inftyint_0^inftyint_0^{exp(-x-y)}dzdxdy.$$
However this evaluates to:
$$V=left[e^yright]_{-infty}^infty.$$
which is clearly divergent.
How should I calculate what the limits are for this volume?
calculus integration definite-integrals
edited Nov 13 at 13:36
Ernie060
2,520319
asked Nov 13 at 13:30
Henry Lee
1,626117
I have a few problems to do with finding the volume under surfaces. The first ones were fairly simple:
$$h(x,y)=xy^2,,0le xle1,,1le yle2$$
$$therefore V=int_1^2int_0^1int_0^{xy^2}dzdxdy=int_1^2int_0^1xy^2dxdy=frac{1}{2}int_1^2y^2dy=frac{7}{6}$$
However, on later questions the boundaries are given as:
$$h(x,y)=e^{-x-y},,0le x,,y<infty$$
which to me implies:
$$V=int_{-infty}^inftyint_0^inftyint_0^{exp(-x-y)}dzdxdy.$$
However this evaluates to:
$$V=left[e^yright]_{-infty}^infty.$$
which is clearly divergent.
How should I calculate what the limits are for this volume?
calculus integration definite-integrals
calculus integration definite-integrals
edited Nov 13 at 13:36
Ernie060
2,520319
asked Nov 13 at 13:30
Henry Lee
1,626117
edited Nov 13 at 13:36
Ernie060
2,520319
asked Nov 13 at 13:30
Henry Lee
1,626117
edited Nov 13 at 13:36
Ernie060
2,520319
edited Nov 13 at 13:36
Ernie060
2,520319
edited Nov 13 at 13:36
Ernie060
2,520319
2,520319
asked Nov 13 at 13:30
Henry Lee
1,626117
asked Nov 13 at 13:30
Henry Lee
1,626117
asked Nov 13 at 13:30
Henry Lee
1,626117
1,626117
add a comment |
add a comment |
1 Answer
1
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oldest
votes
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0
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accepted
The boundaries $0leq x,y < infty$ mean that both $x$ and $y$ can have non-negative values:
$$
begin{align*}
0 leq x < infty \
0 leq y < infty.
end{align*}
$$
So the lower bound in the outer integral (integrating w.r.t $y$) should be $0$.
answered Nov 13 at 13:35
Ernie060
2,520319
Thank you, that makes much more sense now
– Henry Lee
Nov 13 at 13:36
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
The boundaries $0leq x,y < infty$ mean that both $x$ and $y$ can have non-negative values:
$$
begin{align*}
0 leq x < infty \
0 leq y < infty.
end{align*}
$$
So the lower bound in the outer integral (integrating w.r.t $y$) should be $0$.
answered Nov 13 at 13:35
Ernie060
2,520319
Thank you, that makes much more sense now
– Henry Lee
Nov 13 at 13:36
add a comment |
up vote
0
down vote
accepted
The boundaries $0leq x,y < infty$ mean that both $x$ and $y$ can have non-negative values:
$$
begin{align*}
0 leq x < infty \
0 leq y < infty.
end{align*}
$$
So the lower bound in the outer integral (integrating w.r.t $y$) should be $0$.
answered Nov 13 at 13:35
Ernie060
2,520319
Thank you, that makes much more sense now
– Henry Lee
Nov 13 at 13:36
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
The boundaries $0leq x,y < infty$ mean that both $x$ and $y$ can have non-negative values:
$$
begin{align*}
0 leq x < infty \
0 leq y < infty.
end{align*}
$$
So the lower bound in the outer integral (integrating w.r.t $y$) should be $0$.
answered Nov 13 at 13:35
Ernie060
2,520319
The boundaries $0leq x,y < infty$ mean that both $x$ and $y$ can have non-negative values:
$$
begin{align*}
0 leq x < infty \
0 leq y < infty.
end{align*}
$$
So the lower bound in the outer integral (integrating w.r.t $y$) should be $0$.
answered Nov 13 at 13:35
Ernie060
2,520319
edited Nov 13 at 13:38
answered Nov 13 at 13:35
Ernie060
2,520319
answered Nov 13 at 13:35
Ernie060
2,520319
answered Nov 13 at 13:35
Ernie060
2,520319
2,520319
Thank you, that makes much more sense now
– Henry Lee
Nov 13 at 13:36
add a comment |
Thank you, that makes much more sense now
– Henry Lee
Nov 13 at 13:36
Thank you, that makes much more sense now
– Henry Lee
Nov 13 at 13:36
Thank you, that makes much more sense now
– Henry Lee
Nov 13 at 13:36
add a comment |
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