Find the volume of the solid that results when the region enclosed by the given curves is revolved about the...












0















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







share|cite|improve this question
























  • 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
















0















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







share|cite|improve this question
























  • 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














0












0








0








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







share|cite|improve this question
















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 22 '18 at 17:51









Robert Howard

1,9161822




1,9161822










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


















  • 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










2 Answers
2






active

oldest

votes


















1














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.



enter image description here




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.






share|cite|improve this answer



















  • 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














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
$$V=2piint_0^{1/sqrt{5}}xg(x)dx+2piint_{1/sqrt{5}}^1xf(x)dx=2piint_0^{1/sqrt{5}}2xdx+2piint_{1/sqrt{5}}^1xsqrt{frac{1-x^2}{x^2}}dx.$$






share|cite|improve this answer























  • 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













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2 Answers
2






active

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2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1














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.



enter image description here




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.






share|cite|improve this answer



















  • 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














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.



enter image description here




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.






share|cite|improve this answer



















  • 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








1






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.



enter image description here




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.






share|cite|improve this answer














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.



enter image description here




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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 22 '18 at 18:27

























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














  • 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











1














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
$$V=2piint_0^{1/sqrt{5}}xg(x)dx+2piint_{1/sqrt{5}}^1xf(x)dx=2piint_0^{1/sqrt{5}}2xdx+2piint_{1/sqrt{5}}^1xsqrt{frac{1-x^2}{x^2}}dx.$$






share|cite|improve this answer























  • 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


















1














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
$$V=2piint_0^{1/sqrt{5}}xg(x)dx+2piint_{1/sqrt{5}}^1xf(x)dx=2piint_0^{1/sqrt{5}}2xdx+2piint_{1/sqrt{5}}^1xsqrt{frac{1-x^2}{x^2}}dx.$$






share|cite|improve this answer























  • 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
















1












1








1






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
$$V=2piint_0^{1/sqrt{5}}xg(x)dx+2piint_{1/sqrt{5}}^1xf(x)dx=2piint_0^{1/sqrt{5}}2xdx+2piint_{1/sqrt{5}}^1xsqrt{frac{1-x^2}{x^2}}dx.$$






share|cite|improve this answer














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
$$V=2piint_0^{1/sqrt{5}}xg(x)dx+2piint_{1/sqrt{5}}^1xf(x)dx=2piint_0^{1/sqrt{5}}2xdx+2piint_{1/sqrt{5}}^1xsqrt{frac{1-x^2}{x^2}}dx.$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 22 '18 at 17:56

























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




















  • 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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