Find the volume of the solid that results when the region enclosed by the given curves is revolved about the...
Q:Find the volume of the solid that results when the region enclosed by the given curves is revolved about the $y$-axis:
$$y=sqrt{frac{1-x^2}{x^2}}:(x>0),x=0,y=0,y=2$$
My first problem is I can't imagine the region and the answer provided by the book is: $pi tan^{-1}2$ which is far away from what I figured out. Any hints or solution will be appreciated.
Thanks in advance.
calculus solid-of-revolution
edited Nov 22 '18 at 17:51
Robert Howard
1,9161822
asked Nov 22 '18 at 16:44
raihan hossainraihan hossain
868
add a comment |
Q:Find the volume of the solid that results when the region enclosed by the given curves is revolved about the $y$-axis:
$$y=sqrt{frac{1-x^2}{x^2}}:(x>0),x=0,y=0,y=2$$
My first problem is I can't imagine the region and the answer provided by the book is: $pi tan^{-1}2$ which is far away from what I figured out. Any hints or solution will be appreciated.
Thanks in advance.
calculus solid-of-revolution
edited Nov 22 '18 at 17:51
Robert Howard
1,9161822
asked Nov 22 '18 at 16:44
raihan hossainraihan hossain
868
It would help a lot if you showed us your calculations. It's hard to say what you're doing wrong if you don't show us what you've done.
– saulspatz
Nov 22 '18 at 17:13
add a comment |
Q:Find the volume of the solid that results when the region enclosed by the given curves is revolved about the $y$-axis:
$$y=sqrt{frac{1-x^2}{x^2}}:(x>0),x=0,y=0,y=2$$
My first problem is I can't imagine the region and the answer provided by the book is: $pi tan^{-1}2$ which is far away from what I figured out. Any hints or solution will be appreciated.
Thanks in advance.
calculus solid-of-revolution
edited Nov 22 '18 at 17:51
Robert Howard
1,9161822
asked Nov 22 '18 at 16:44
raihan hossainraihan hossain
868
Q:Find the volume of the solid that results when the region enclosed by the given curves is revolved about the $y$-axis:
$$y=sqrt{frac{1-x^2}{x^2}}:(x>0),x=0,y=0,y=2$$
My first problem is I can't imagine the region and the answer provided by the book is: $pi tan^{-1}2$ which is far away from what I figured out. Any hints or solution will be appreciated.
Thanks in advance.
calculus solid-of-revolution
calculus solid-of-revolution
edited Nov 22 '18 at 17:51
Robert Howard
1,9161822
asked Nov 22 '18 at 16:44
raihan hossainraihan hossain
868
edited Nov 22 '18 at 17:51
Robert Howard
1,9161822
asked Nov 22 '18 at 16:44
raihan hossainraihan hossain
868
edited Nov 22 '18 at 17:51
Robert Howard
1,9161822
edited Nov 22 '18 at 17:51
Robert Howard
1,9161822
edited Nov 22 '18 at 17:51
Robert Howard
1,9161822
1,9161822
asked Nov 22 '18 at 16:44
raihan hossainraihan hossain
868
asked Nov 22 '18 at 16:44
raihan hossainraihan hossain
868
asked Nov 22 '18 at 16:44
raihan hossainraihan hossain
868
868
It would help a lot if you showed us your calculations. It's hard to say what you're doing wrong if you don't show us what you've done.
– saulspatz
Nov 22 '18 at 17:13
add a comment |
It would help a lot if you showed us your calculations. It's hard to say what you're doing wrong if you don't show us what you've done.
– saulspatz
Nov 22 '18 at 17:13
It would help a lot if you showed us your calculations. It's hard to say what you're doing wrong if you don't show us what you've done.
– saulspatz
Nov 22 '18 at 17:13
It would help a lot if you showed us your calculations. It's hard to say what you're doing wrong if you don't show us what you've done.
– saulspatz
Nov 22 '18 at 17:13
add a comment |
2 Answers
2
active
oldest
votes
Here's the graph of the function (in red) with the area to be revolved around the $y$-axis shaded in blue.
There are two ways you could do this: by re-expressing the function in terms of $y$ and using the disc method, or by leaving the function the way it is and using the shell method.
Using the first method, you would need to evaluate the integral $$piint_0^2[g(y)]^2dy,$$ where $g(y)$ is the same function, but expressed in terms of $y$ (in other words, $x=ldots$ instead of $y=ldots$).
To find the volume using the second method, you would need two integrals (can you tell why?): $$2piint_0^ax(2)dx+2piint_a^1xf(x)dx,$$ where $x=a$ is the point of intersection between the function and the line $y=2$, which I'll leave to you to calculate.
Here's a brief explanation of why I used two integrals in the second method:
The general formula for finding the volume of a solid of revolution with the shell method is $$2piint_a^bxf(x),dx,$$ where $f(x)$ is some function that gives the height of the region you want to revolve around the $y$-axis. In this case, from $x=0$ to $x=1/sqrt{5}$, that function is just $y=2$, while from $x=1/sqrt{5}$ to $x=1$, the height is given by $y=sqrt{frac{1-x^2}{x^2}}$. Because the two regions have different heights, we need to use separate integrals to find the volume obtained from revolving each region around the $y$-axis.
In general, any time you see a sharp corner like the one at $(1/sqrt{5},2)$, that's a sign that you'll need multiple integrals.
answered Nov 22 '18 at 17:36
Robert HowardRobert Howard
1,9161822
1
i need two integrals because you divides the volume into two cylinder at $frac {1}{ sqrt 5}$.Am i right?
– raihan hossain
Nov 22 '18 at 18:10
From a geometrical standpoint, that's correct. Another way of asking the same question would be this: why would the integral $$2piint_0^1xf(x)dx$$ not give you the right volume?
– Robert Howard
Nov 22 '18 at 18:13
Really why? Can you explain it for me @Robert Howard Sir
– raihan hossain
Nov 22 '18 at 18:18
Sure; I'll add an explanation to the end of my answer.
– Robert Howard
Nov 22 '18 at 18:22
Thanks a lot @Robert Howard Sir.Now I understand this question as well as this method :)
– raihan hossain
Nov 22 '18 at 18:41
|
show 1 more comment
The formula to compute volumes revolved about the $y$-axis is
$$V=2piint_a^bxf(x)dx.$$
First check where $f(x)=2$:
$$sqrt{frac{1-x^2}{x^2}}=2implies 1-x^2=4x^2implies x=frac{1}{sqrt{5}}.$$
So you can integrate between $0$ and $1/sqrt{5}$ with $g(x)=2$ and the rest of the area from $1/sqrt{5}$ to $1$ with your function $f(x)$, you get
$$V=2piint_0^{1/sqrt{5}}xg(x)dx+2piint_{1/sqrt{5}}^1xf(x)dx=2piint_0^{1/sqrt{5}}2xdx+2piint_{1/sqrt{5}}^1xsqrt{frac{1-x^2}{x^2}}dx.$$
answered Nov 22 '18 at 17:40
james wattjames watt
34610
Why $V=int_a^b pi f(y)^2 dy$ this formula is not used by you @james watt Sir.
– raihan hossain
Nov 22 '18 at 17:53
That formula is used to calculate volumes around the $x$-axis. If you rotate your function around $x$-axis the value $f(x)$ is the radius of your solid in a certain $xin[a,b]$, integrating on $[a,b]$ you get the total volume. $f(x)$ is squared because the formula of the area of a circle is $pi r^2$, in this case $r=f(x)$.
– james watt
Nov 22 '18 at 17:57
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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oldest
votes
Here's the graph of the function (in red) with the area to be revolved around the $y$-axis shaded in blue.
There are two ways you could do this: by re-expressing the function in terms of $y$ and using the disc method, or by leaving the function the way it is and using the shell method.
Using the first method, you would need to evaluate the integral $$piint_0^2[g(y)]^2dy,$$ where $g(y)$ is the same function, but expressed in terms of $y$ (in other words, $x=ldots$ instead of $y=ldots$).
To find the volume using the second method, you would need two integrals (can you tell why?): $$2piint_0^ax(2)dx+2piint_a^1xf(x)dx,$$ where $x=a$ is the point of intersection between the function and the line $y=2$, which I'll leave to you to calculate.
Here's a brief explanation of why I used two integrals in the second method:
The general formula for finding the volume of a solid of revolution with the shell method is $$2piint_a^bxf(x),dx,$$ where $f(x)$ is some function that gives the height of the region you want to revolve around the $y$-axis. In this case, from $x=0$ to $x=1/sqrt{5}$, that function is just $y=2$, while from $x=1/sqrt{5}$ to $x=1$, the height is given by $y=sqrt{frac{1-x^2}{x^2}}$. Because the two regions have different heights, we need to use separate integrals to find the volume obtained from revolving each region around the $y$-axis.
In general, any time you see a sharp corner like the one at $(1/sqrt{5},2)$, that's a sign that you'll need multiple integrals.
answered Nov 22 '18 at 17:36
Robert HowardRobert Howard
1,9161822
1
i need two integrals because you divides the volume into two cylinder at $frac {1}{ sqrt 5}$.Am i right?
– raihan hossain
Nov 22 '18 at 18:10
From a geometrical standpoint, that's correct. Another way of asking the same question would be this: why would the integral $$2piint_0^1xf(x)dx$$ not give you the right volume?
– Robert Howard
Nov 22 '18 at 18:13
Really why? Can you explain it for me @Robert Howard Sir
– raihan hossain
Nov 22 '18 at 18:18
Sure; I'll add an explanation to the end of my answer.
– Robert Howard
Nov 22 '18 at 18:22
Thanks a lot @Robert Howard Sir.Now I understand this question as well as this method :)
– raihan hossain
Nov 22 '18 at 18:41
|
show 1 more comment
Here's the graph of the function (in red) with the area to be revolved around the $y$-axis shaded in blue.
There are two ways you could do this: by re-expressing the function in terms of $y$ and using the disc method, or by leaving the function the way it is and using the shell method.
Using the first method, you would need to evaluate the integral $$piint_0^2[g(y)]^2dy,$$ where $g(y)$ is the same function, but expressed in terms of $y$ (in other words, $x=ldots$ instead of $y=ldots$).
To find the volume using the second method, you would need two integrals (can you tell why?): $$2piint_0^ax(2)dx+2piint_a^1xf(x)dx,$$ where $x=a$ is the point of intersection between the function and the line $y=2$, which I'll leave to you to calculate.
Here's a brief explanation of why I used two integrals in the second method:
The general formula for finding the volume of a solid of revolution with the shell method is $$2piint_a^bxf(x),dx,$$ where $f(x)$ is some function that gives the height of the region you want to revolve around the $y$-axis. In this case, from $x=0$ to $x=1/sqrt{5}$, that function is just $y=2$, while from $x=1/sqrt{5}$ to $x=1$, the height is given by $y=sqrt{frac{1-x^2}{x^2}}$. Because the two regions have different heights, we need to use separate integrals to find the volume obtained from revolving each region around the $y$-axis.
In general, any time you see a sharp corner like the one at $(1/sqrt{5},2)$, that's a sign that you'll need multiple integrals.
answered Nov 22 '18 at 17:36
Robert HowardRobert Howard
1,9161822
1
i need two integrals because you divides the volume into two cylinder at $frac {1}{ sqrt 5}$.Am i right?
– raihan hossain
Nov 22 '18 at 18:10
From a geometrical standpoint, that's correct. Another way of asking the same question would be this: why would the integral $$2piint_0^1xf(x)dx$$ not give you the right volume?
– Robert Howard
Nov 22 '18 at 18:13
Really why? Can you explain it for me @Robert Howard Sir
– raihan hossain
Nov 22 '18 at 18:18
Sure; I'll add an explanation to the end of my answer.
– Robert Howard
Nov 22 '18 at 18:22
Thanks a lot @Robert Howard Sir.Now I understand this question as well as this method :)
– raihan hossain
Nov 22 '18 at 18:41
|
show 1 more comment
Here's the graph of the function (in red) with the area to be revolved around the $y$-axis shaded in blue.
There are two ways you could do this: by re-expressing the function in terms of $y$ and using the disc method, or by leaving the function the way it is and using the shell method.
Using the first method, you would need to evaluate the integral $$piint_0^2[g(y)]^2dy,$$ where $g(y)$ is the same function, but expressed in terms of $y$ (in other words, $x=ldots$ instead of $y=ldots$).
To find the volume using the second method, you would need two integrals (can you tell why?): $$2piint_0^ax(2)dx+2piint_a^1xf(x)dx,$$ where $x=a$ is the point of intersection between the function and the line $y=2$, which I'll leave to you to calculate.
Here's a brief explanation of why I used two integrals in the second method:
The general formula for finding the volume of a solid of revolution with the shell method is $$2piint_a^bxf(x),dx,$$ where $f(x)$ is some function that gives the height of the region you want to revolve around the $y$-axis. In this case, from $x=0$ to $x=1/sqrt{5}$, that function is just $y=2$, while from $x=1/sqrt{5}$ to $x=1$, the height is given by $y=sqrt{frac{1-x^2}{x^2}}$. Because the two regions have different heights, we need to use separate integrals to find the volume obtained from revolving each region around the $y$-axis.
In general, any time you see a sharp corner like the one at $(1/sqrt{5},2)$, that's a sign that you'll need multiple integrals.
answered Nov 22 '18 at 17:36
Robert HowardRobert Howard
1,9161822
Here's the graph of the function (in red) with the area to be revolved around the $y$-axis shaded in blue.
There are two ways you could do this: by re-expressing the function in terms of $y$ and using the disc method, or by leaving the function the way it is and using the shell method.
Using the first method, you would need to evaluate the integral $$piint_0^2[g(y)]^2dy,$$ where $g(y)$ is the same function, but expressed in terms of $y$ (in other words, $x=ldots$ instead of $y=ldots$).
To find the volume using the second method, you would need two integrals (can you tell why?): $$2piint_0^ax(2)dx+2piint_a^1xf(x)dx,$$ where $x=a$ is the point of intersection between the function and the line $y=2$, which I'll leave to you to calculate.
Here's a brief explanation of why I used two integrals in the second method:
The general formula for finding the volume of a solid of revolution with the shell method is $$2piint_a^bxf(x),dx,$$ where $f(x)$ is some function that gives the height of the region you want to revolve around the $y$-axis. In this case, from $x=0$ to $x=1/sqrt{5}$, that function is just $y=2$, while from $x=1/sqrt{5}$ to $x=1$, the height is given by $y=sqrt{frac{1-x^2}{x^2}}$. Because the two regions have different heights, we need to use separate integrals to find the volume obtained from revolving each region around the $y$-axis.
In general, any time you see a sharp corner like the one at $(1/sqrt{5},2)$, that's a sign that you'll need multiple integrals.
answered Nov 22 '18 at 17:36
Robert HowardRobert Howard
1,9161822
edited Nov 22 '18 at 18:27
answered Nov 22 '18 at 17:36
Robert HowardRobert Howard
1,9161822
answered Nov 22 '18 at 17:36
Robert HowardRobert Howard
1,9161822
answered Nov 22 '18 at 17:36
Robert HowardRobert Howard
1,9161822
1,9161822
1
i need two integrals because you divides the volume into two cylinder at $frac {1}{ sqrt 5}$.Am i right?
– raihan hossain
Nov 22 '18 at 18:10
From a geometrical standpoint, that's correct. Another way of asking the same question would be this: why would the integral $$2piint_0^1xf(x)dx$$ not give you the right volume?
– Robert Howard
Nov 22 '18 at 18:13
Really why? Can you explain it for me @Robert Howard Sir
– raihan hossain
Nov 22 '18 at 18:18
Sure; I'll add an explanation to the end of my answer.
– Robert Howard
Nov 22 '18 at 18:22
Thanks a lot @Robert Howard Sir.Now I understand this question as well as this method :)
– raihan hossain
Nov 22 '18 at 18:41
|
show 1 more comment
1
i need two integrals because you divides the volume into two cylinder at $frac {1}{ sqrt 5}$.Am i right?
– raihan hossain
Nov 22 '18 at 18:10
From a geometrical standpoint, that's correct. Another way of asking the same question would be this: why would the integral $$2piint_0^1xf(x)dx$$ not give you the right volume?
– Robert Howard
Nov 22 '18 at 18:13
Really why? Can you explain it for me @Robert Howard Sir
– raihan hossain
Nov 22 '18 at 18:18
Sure; I'll add an explanation to the end of my answer.
– Robert Howard
Nov 22 '18 at 18:22
Thanks a lot @Robert Howard Sir.Now I understand this question as well as this method :)
– raihan hossain
Nov 22 '18 at 18:41
1
1
i need two integrals because you divides the volume into two cylinder at $frac {1}{ sqrt 5}$.Am i right?
– raihan hossain
Nov 22 '18 at 18:10
i need two integrals because you divides the volume into two cylinder at $frac {1}{ sqrt 5}$.Am i right?
– raihan hossain
Nov 22 '18 at 18:10
From a geometrical standpoint, that's correct. Another way of asking the same question would be this: why would the integral $$2piint_0^1xf(x)dx$$ not give you the right volume?
– Robert Howard
Nov 22 '18 at 18:13
From a geometrical standpoint, that's correct. Another way of asking the same question would be this: why would the integral $$2piint_0^1xf(x)dx$$ not give you the right volume?
– Robert Howard
Nov 22 '18 at 18:13
Really why? Can you explain it for me @Robert Howard Sir
– raihan hossain
Nov 22 '18 at 18:18
Really why? Can you explain it for me @Robert Howard Sir
– raihan hossain
Nov 22 '18 at 18:18
Sure; I'll add an explanation to the end of my answer.
– Robert Howard
Nov 22 '18 at 18:22
Sure; I'll add an explanation to the end of my answer.
– Robert Howard
Nov 22 '18 at 18:22
Thanks a lot @Robert Howard Sir.Now I understand this question as well as this method :)
– raihan hossain
Nov 22 '18 at 18:41
Thanks a lot @Robert Howard Sir.Now I understand this question as well as this method :)
– raihan hossain
Nov 22 '18 at 18:41
|
show 1 more comment
The formula to compute volumes revolved about the $y$-axis is
$$V=2piint_a^bxf(x)dx.$$
First check where $f(x)=2$:
$$sqrt{frac{1-x^2}{x^2}}=2implies 1-x^2=4x^2implies x=frac{1}{sqrt{5}}.$$
So you can integrate between $0$ and $1/sqrt{5}$ with $g(x)=2$ and the rest of the area from $1/sqrt{5}$ to $1$ with your function $f(x)$, you get
$$V=2piint_0^{1/sqrt{5}}xg(x)dx+2piint_{1/sqrt{5}}^1xf(x)dx=2piint_0^{1/sqrt{5}}2xdx+2piint_{1/sqrt{5}}^1xsqrt{frac{1-x^2}{x^2}}dx.$$
answered Nov 22 '18 at 17:40
james wattjames watt
34610
Why $V=int_a^b pi f(y)^2 dy$ this formula is not used by you @james watt Sir.
– raihan hossain
Nov 22 '18 at 17:53
That formula is used to calculate volumes around the $x$-axis. If you rotate your function around $x$-axis the value $f(x)$ is the radius of your solid in a certain $xin[a,b]$, integrating on $[a,b]$ you get the total volume. $f(x)$ is squared because the formula of the area of a circle is $pi r^2$, in this case $r=f(x)$.
– james watt
Nov 22 '18 at 17:57
add a comment |
The formula to compute volumes revolved about the $y$-axis is
$$V=2piint_a^bxf(x)dx.$$
First check where $f(x)=2$:
$$sqrt{frac{1-x^2}{x^2}}=2implies 1-x^2=4x^2implies x=frac{1}{sqrt{5}}.$$
So you can integrate between $0$ and $1/sqrt{5}$ with $g(x)=2$ and the rest of the area from $1/sqrt{5}$ to $1$ with your function $f(x)$, you get
$$V=2piint_0^{1/sqrt{5}}xg(x)dx+2piint_{1/sqrt{5}}^1xf(x)dx=2piint_0^{1/sqrt{5}}2xdx+2piint_{1/sqrt{5}}^1xsqrt{frac{1-x^2}{x^2}}dx.$$
answered Nov 22 '18 at 17:40
james wattjames watt
34610
Why $V=int_a^b pi f(y)^2 dy$ this formula is not used by you @james watt Sir.
– raihan hossain
Nov 22 '18 at 17:53
That formula is used to calculate volumes around the $x$-axis. If you rotate your function around $x$-axis the value $f(x)$ is the radius of your solid in a certain $xin[a,b]$, integrating on $[a,b]$ you get the total volume. $f(x)$ is squared because the formula of the area of a circle is $pi r^2$, in this case $r=f(x)$.
– james watt
Nov 22 '18 at 17:57
add a comment |
The formula to compute volumes revolved about the $y$-axis is
$$V=2piint_a^bxf(x)dx.$$
First check where $f(x)=2$:
$$sqrt{frac{1-x^2}{x^2}}=2implies 1-x^2=4x^2implies x=frac{1}{sqrt{5}}.$$
So you can integrate between $0$ and $1/sqrt{5}$ with $g(x)=2$ and the rest of the area from $1/sqrt{5}$ to $1$ with your function $f(x)$, you get
$$V=2piint_0^{1/sqrt{5}}xg(x)dx+2piint_{1/sqrt{5}}^1xf(x)dx=2piint_0^{1/sqrt{5}}2xdx+2piint_{1/sqrt{5}}^1xsqrt{frac{1-x^2}{x^2}}dx.$$
answered Nov 22 '18 at 17:40
james wattjames watt
34610
The formula to compute volumes revolved about the $y$-axis is
$$V=2piint_a^bxf(x)dx.$$
First check where $f(x)=2$:
$$sqrt{frac{1-x^2}{x^2}}=2implies 1-x^2=4x^2implies x=frac{1}{sqrt{5}}.$$
So you can integrate between $0$ and $1/sqrt{5}$ with $g(x)=2$ and the rest of the area from $1/sqrt{5}$ to $1$ with your function $f(x)$, you get
$$V=2piint_0^{1/sqrt{5}}xg(x)dx+2piint_{1/sqrt{5}}^1xf(x)dx=2piint_0^{1/sqrt{5}}2xdx+2piint_{1/sqrt{5}}^1xsqrt{frac{1-x^2}{x^2}}dx.$$
answered Nov 22 '18 at 17:40
james wattjames watt
34610
edited Nov 22 '18 at 17:56
answered Nov 22 '18 at 17:40
james wattjames watt
34610
answered Nov 22 '18 at 17:40
james wattjames watt
34610
answered Nov 22 '18 at 17:40
james wattjames watt
34610
34610
Why $V=int_a^b pi f(y)^2 dy$ this formula is not used by you @james watt Sir.
– raihan hossain
Nov 22 '18 at 17:53
That formula is used to calculate volumes around the $x$-axis. If you rotate your function around $x$-axis the value $f(x)$ is the radius of your solid in a certain $xin[a,b]$, integrating on $[a,b]$ you get the total volume. $f(x)$ is squared because the formula of the area of a circle is $pi r^2$, in this case $r=f(x)$.
– james watt
Nov 22 '18 at 17:57
add a comment |
Why $V=int_a^b pi f(y)^2 dy$ this formula is not used by you @james watt Sir.
– raihan hossain
Nov 22 '18 at 17:53
That formula is used to calculate volumes around the $x$-axis. If you rotate your function around $x$-axis the value $f(x)$ is the radius of your solid in a certain $xin[a,b]$, integrating on $[a,b]$ you get the total volume. $f(x)$ is squared because the formula of the area of a circle is $pi r^2$, in this case $r=f(x)$.
– james watt
Nov 22 '18 at 17:57
Why $V=int_a^b pi f(y)^2 dy$ this formula is not used by you @james watt Sir.
– raihan hossain
Nov 22 '18 at 17:53
Why $V=int_a^b pi f(y)^2 dy$ this formula is not used by you @james watt Sir.
– raihan hossain
Nov 22 '18 at 17:53
That formula is used to calculate volumes around the $x$-axis. If you rotate your function around $x$-axis the value $f(x)$ is the radius of your solid in a certain $xin[a,b]$, integrating on $[a,b]$ you get the total volume. $f(x)$ is squared because the formula of the area of a circle is $pi r^2$, in this case $r=f(x)$.
– james watt
Nov 22 '18 at 17:57
That formula is used to calculate volumes around the $x$-axis. If you rotate your function around $x$-axis the value $f(x)$ is the radius of your solid in a certain $xin[a,b]$, integrating on $[a,b]$ you get the total volume. $f(x)$ is squared because the formula of the area of a circle is $pi r^2$, in this case $r=f(x)$.
– james watt
Nov 22 '18 at 17:57
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It would help a lot if you showed us your calculations. It's hard to say what you're doing wrong if you don't show us what you've done.
– saulspatz
Nov 22 '18 at 17:13