Compute the volume of a solid by revolving a region about the $y$-axis
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How to compute the volume of the solid generated by revolving the region between the curve $y=dfrac{cos x}{x}$ and the $x$-axis for $pi/6leq xleq pi/2$ about the $y$-axis?
I think we need to compute a definite integral of the form $int_a^b pi x^2dy$, however it's impossible to express $x$ in terms of $y$ explicitly.
Can someone help me? Thanks a lot!
calculus integration volume
asked Dec 24 '18 at 5:05
mappingmapping
1778
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add a comment |
$begingroup$
How to compute the volume of the solid generated by revolving the region between the curve $y=dfrac{cos x}{x}$ and the $x$-axis for $pi/6leq xleq pi/2$ about the $y$-axis?
I think we need to compute a definite integral of the form $int_a^b pi x^2dy$, however it's impossible to express $x$ in terms of $y$ explicitly.
Can someone help me? Thanks a lot!
calculus integration volume
asked Dec 24 '18 at 5:05
mappingmapping
1778
$endgroup$
add a comment |
$begingroup$
How to compute the volume of the solid generated by revolving the region between the curve $y=dfrac{cos x}{x}$ and the $x$-axis for $pi/6leq xleq pi/2$ about the $y$-axis?
I think we need to compute a definite integral of the form $int_a^b pi x^2dy$, however it's impossible to express $x$ in terms of $y$ explicitly.
Can someone help me? Thanks a lot!
calculus integration volume
asked Dec 24 '18 at 5:05
mappingmapping
1778
$endgroup$
How to compute the volume of the solid generated by revolving the region between the curve $y=dfrac{cos x}{x}$ and the $x$-axis for $pi/6leq xleq pi/2$ about the $y$-axis?
I think we need to compute a definite integral of the form $int_a^b pi x^2dy$, however it's impossible to express $x$ in terms of $y$ explicitly.
Can someone help me? Thanks a lot!
calculus integration volume
calculus integration volume
asked Dec 24 '18 at 5:05
mappingmapping
1778
asked Dec 24 '18 at 5:05
mappingmapping
1778
asked Dec 24 '18 at 5:05
mappingmapping
1778
asked Dec 24 '18 at 5:05
mappingmapping
1778
asked Dec 24 '18 at 5:05
mappingmapping
1778
1778
add a comment |
add a comment |
3 Answers
3
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oldest
votes
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The shell method allows you to calculate the volume of a solid of revolution that rotates to an axis that is perpendicular to the one where we're integrating - in this case, we can rotate a shape around the $y$-axis by integrating over $x$. So you don't need to express the function in terms of $y$.
The volume of the solid will be an integral of the form
$$V=2pi int_a^b x|f(x)|dx,$$
where in this case $f(x)=frac{cos(x)}{x}$, $a=pi/6$ and $b=pi/2$. Note that $f(x)geq 0$ in the interval $[frac pi 6, frac pi 2]$ (check this yourself). Thus we get:
$$V=2pi int_{pi/6}^{pi/2} xleft|frac{cos(x)}{x}right|dx=2pi int_{pi/6}^{pi/2} xfrac{cos(x)}{x}dx=2piint_{pi/6}^{pi/2} cos(x)dx=2pi(sin(x))^{pi/2}_{pi/6}=2pi(1-frac 1 2)=pi.$$
answered Dec 24 '18 at 5:38
Bastián NúñezBastián Núñez
1765
$endgroup$
add a comment |
$begingroup$
You may use shell method to compute the volume. Then volume is $$int_{pi/6}^{pi/2}2pi xfrac{cos x}{x}dx=2piint_{pi/6}^{pi/2}cos x dx=pi$$
answered Dec 24 '18 at 5:26
guestguest
775416
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add a comment |
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There is another formula, namely, $$int 2pi x f(x) dx $$ which could be used to find the volume of revolution about the y-axis.
You will get $$int_{pi/6}^{pi/2} 2pi cos x dx =pi$$
answered Dec 24 '18 at 5:26
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The shell method allows you to calculate the volume of a solid of revolution that rotates to an axis that is perpendicular to the one where we're integrating - in this case, we can rotate a shape around the $y$-axis by integrating over $x$. So you don't need to express the function in terms of $y$.
The volume of the solid will be an integral of the form
$$V=2pi int_a^b x|f(x)|dx,$$
where in this case $f(x)=frac{cos(x)}{x}$, $a=pi/6$ and $b=pi/2$. Note that $f(x)geq 0$ in the interval $[frac pi 6, frac pi 2]$ (check this yourself). Thus we get:
$$V=2pi int_{pi/6}^{pi/2} xleft|frac{cos(x)}{x}right|dx=2pi int_{pi/6}^{pi/2} xfrac{cos(x)}{x}dx=2piint_{pi/6}^{pi/2} cos(x)dx=2pi(sin(x))^{pi/2}_{pi/6}=2pi(1-frac 1 2)=pi.$$
answered Dec 24 '18 at 5:38
Bastián NúñezBastián Núñez
1765
$endgroup$
add a comment |
$begingroup$
The shell method allows you to calculate the volume of a solid of revolution that rotates to an axis that is perpendicular to the one where we're integrating - in this case, we can rotate a shape around the $y$-axis by integrating over $x$. So you don't need to express the function in terms of $y$.
The volume of the solid will be an integral of the form
$$V=2pi int_a^b x|f(x)|dx,$$
where in this case $f(x)=frac{cos(x)}{x}$, $a=pi/6$ and $b=pi/2$. Note that $f(x)geq 0$ in the interval $[frac pi 6, frac pi 2]$ (check this yourself). Thus we get:
$$V=2pi int_{pi/6}^{pi/2} xleft|frac{cos(x)}{x}right|dx=2pi int_{pi/6}^{pi/2} xfrac{cos(x)}{x}dx=2piint_{pi/6}^{pi/2} cos(x)dx=2pi(sin(x))^{pi/2}_{pi/6}=2pi(1-frac 1 2)=pi.$$
answered Dec 24 '18 at 5:38
Bastián NúñezBastián Núñez
1765
$endgroup$
add a comment |
$begingroup$
The shell method allows you to calculate the volume of a solid of revolution that rotates to an axis that is perpendicular to the one where we're integrating - in this case, we can rotate a shape around the $y$-axis by integrating over $x$. So you don't need to express the function in terms of $y$.
The volume of the solid will be an integral of the form
$$V=2pi int_a^b x|f(x)|dx,$$
where in this case $f(x)=frac{cos(x)}{x}$, $a=pi/6$ and $b=pi/2$. Note that $f(x)geq 0$ in the interval $[frac pi 6, frac pi 2]$ (check this yourself). Thus we get:
$$V=2pi int_{pi/6}^{pi/2} xleft|frac{cos(x)}{x}right|dx=2pi int_{pi/6}^{pi/2} xfrac{cos(x)}{x}dx=2piint_{pi/6}^{pi/2} cos(x)dx=2pi(sin(x))^{pi/2}_{pi/6}=2pi(1-frac 1 2)=pi.$$
answered Dec 24 '18 at 5:38
Bastián NúñezBastián Núñez
1765
$endgroup$
The shell method allows you to calculate the volume of a solid of revolution that rotates to an axis that is perpendicular to the one where we're integrating - in this case, we can rotate a shape around the $y$-axis by integrating over $x$. So you don't need to express the function in terms of $y$.
The volume of the solid will be an integral of the form
$$V=2pi int_a^b x|f(x)|dx,$$
where in this case $f(x)=frac{cos(x)}{x}$, $a=pi/6$ and $b=pi/2$. Note that $f(x)geq 0$ in the interval $[frac pi 6, frac pi 2]$ (check this yourself). Thus we get:
$$V=2pi int_{pi/6}^{pi/2} xleft|frac{cos(x)}{x}right|dx=2pi int_{pi/6}^{pi/2} xfrac{cos(x)}{x}dx=2piint_{pi/6}^{pi/2} cos(x)dx=2pi(sin(x))^{pi/2}_{pi/6}=2pi(1-frac 1 2)=pi.$$
answered Dec 24 '18 at 5:38
Bastián NúñezBastián Núñez
1765
answered Dec 24 '18 at 5:38
Bastián NúñezBastián Núñez
1765
answered Dec 24 '18 at 5:38
Bastián NúñezBastián Núñez
1765
answered Dec 24 '18 at 5:38
Bastián NúñezBastián Núñez
1765
1765
add a comment |
add a comment |
$begingroup$
You may use shell method to compute the volume. Then volume is $$int_{pi/6}^{pi/2}2pi xfrac{cos x}{x}dx=2piint_{pi/6}^{pi/2}cos x dx=pi$$
answered Dec 24 '18 at 5:26
guestguest
775416
$endgroup$
add a comment |
$begingroup$
You may use shell method to compute the volume. Then volume is $$int_{pi/6}^{pi/2}2pi xfrac{cos x}{x}dx=2piint_{pi/6}^{pi/2}cos x dx=pi$$
answered Dec 24 '18 at 5:26
guestguest
775416
$endgroup$
add a comment |
$begingroup$
You may use shell method to compute the volume. Then volume is $$int_{pi/6}^{pi/2}2pi xfrac{cos x}{x}dx=2piint_{pi/6}^{pi/2}cos x dx=pi$$
answered Dec 24 '18 at 5:26
guestguest
775416
$endgroup$
You may use shell method to compute the volume. Then volume is $$int_{pi/6}^{pi/2}2pi xfrac{cos x}{x}dx=2piint_{pi/6}^{pi/2}cos x dx=pi$$
answered Dec 24 '18 at 5:26
guestguest
775416
edited Dec 24 '18 at 7:49
answered Dec 24 '18 at 5:26
guestguest
775416
answered Dec 24 '18 at 5:26
guestguest
775416
answered Dec 24 '18 at 5:26
guestguest
775416
775416
add a comment |
add a comment |
$begingroup$
There is another formula, namely, $$int 2pi x f(x) dx $$ which could be used to find the volume of revolution about the y-axis.
You will get $$int_{pi/6}^{pi/2} 2pi cos x dx =pi$$
answered Dec 24 '18 at 5:26
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
add a comment |
$begingroup$
There is another formula, namely, $$int 2pi x f(x) dx $$ which could be used to find the volume of revolution about the y-axis.
You will get $$int_{pi/6}^{pi/2} 2pi cos x dx =pi$$
answered Dec 24 '18 at 5:26
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
add a comment |
$begingroup$
There is another formula, namely, $$int 2pi x f(x) dx $$ which could be used to find the volume of revolution about the y-axis.
You will get $$int_{pi/6}^{pi/2} 2pi cos x dx =pi$$
answered Dec 24 '18 at 5:26
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
There is another formula, namely, $$int 2pi x f(x) dx $$ which could be used to find the volume of revolution about the y-axis.
You will get $$int_{pi/6}^{pi/2} 2pi cos x dx =pi$$
answered Dec 24 '18 at 5:26
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
edited Dec 25 '18 at 14:43
answered Dec 24 '18 at 5:26
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
answered Dec 24 '18 at 5:26
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
answered Dec 24 '18 at 5:26
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
41.6k42061
add a comment |
add a comment |
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