Volume of a curve using integration [closed] Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Disk method to find volume of solid revolved around x axisA volume integral questionIntegral volume questionsFind Volume of a solid when the region is revolved about y-axis or x-axisFinding volume of solid using integration.Application of integration. (Volume)Volume through integrationFind volume of a solidRevolving a function around $y$-axis, finding volume using shell methodVolume when the region bounded by $x=y^2$ and $x=y+2$ is revolved about the $y$-axis
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Volume of a curve using integration [closed]
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Disk method to find volume of solid revolved around x axisA volume integral questionIntegral volume questionsFind Volume of a solid when the region is revolved about y-axis or x-axisFinding volume of solid using integration.Application of integration. (Volume)Volume through integrationFind volume of a solidRevolving a function around $y$-axis, finding volume using shell methodVolume when the region bounded by $x=y^2$ and $x=y+2$ is revolved about the $y$-axis
$begingroup$
Consider the region $R$ bounded by the curves $y=ax^2+1, y=0, x=0,spacetextandspace x=1, textforspace ageq-1$.
If $V_1(a)$ is the volume of the solid generated when $R$ is revolved about the $x$-axis and $V_2(a)$ is the volume of the solid generated when $R$ is revolved about $y$-axis.
Find the values of $ageq-1$ for which $V_1(a)=V_2(a)$.
integration volume solid-of-revolution
edited Mar 31 at 19:55
coreyman317
1,280522
asked Mar 31 at 17:56
Ceren BülbülCeren Bülbül
54
$endgroup$
closed as off-topic by Eevee Trainer, José Carlos Santos, John Omielan, Martin Argerami, Tianlalu Apr 1 at 4:12
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, José Carlos Santos, John Omielan, Martin Argerami, Tianlalu
add a comment |
$begingroup$
Consider the region $R$ bounded by the curves $y=ax^2+1, y=0, x=0,spacetextandspace x=1, textforspace ageq-1$.
If $V_1(a)$ is the volume of the solid generated when $R$ is revolved about the $x$-axis and $V_2(a)$ is the volume of the solid generated when $R$ is revolved about $y$-axis.
Find the values of $ageq-1$ for which $V_1(a)=V_2(a)$.
integration volume solid-of-revolution
edited Mar 31 at 19:55
coreyman317
1,280522
asked Mar 31 at 17:56
Ceren BülbülCeren Bülbül
54
$endgroup$
closed as off-topic by Eevee Trainer, José Carlos Santos, John Omielan, Martin Argerami, Tianlalu Apr 1 at 4:12
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, José Carlos Santos, John Omielan, Martin Argerami, Tianlalu
add a comment |
$begingroup$
Consider the region $R$ bounded by the curves $y=ax^2+1, y=0, x=0,spacetextandspace x=1, textforspace ageq-1$.
If $V_1(a)$ is the volume of the solid generated when $R$ is revolved about the $x$-axis and $V_2(a)$ is the volume of the solid generated when $R$ is revolved about $y$-axis.
Find the values of $ageq-1$ for which $V_1(a)=V_2(a)$.
integration volume solid-of-revolution
edited Mar 31 at 19:55
coreyman317
1,280522
asked Mar 31 at 17:56
Ceren BülbülCeren Bülbül
54
$endgroup$
Consider the region $R$ bounded by the curves $y=ax^2+1, y=0, x=0,spacetextandspace x=1, textforspace ageq-1$.
If $V_1(a)$ is the volume of the solid generated when $R$ is revolved about the $x$-axis and $V_2(a)$ is the volume of the solid generated when $R$ is revolved about $y$-axis.
Find the values of $ageq-1$ for which $V_1(a)=V_2(a)$.
integration volume solid-of-revolution
integration volume solid-of-revolution
edited Mar 31 at 19:55
coreyman317
1,280522
asked Mar 31 at 17:56
Ceren BülbülCeren Bülbül
54
edited Mar 31 at 19:55
coreyman317
1,280522
asked Mar 31 at 17:56
Ceren BülbülCeren Bülbül
54
edited Mar 31 at 19:55
coreyman317
1,280522
edited Mar 31 at 19:55
coreyman317
1,280522
edited Mar 31 at 19:55
coreyman317
1,280522
1,280522
asked Mar 31 at 17:56
Ceren BülbülCeren Bülbül
54
asked Mar 31 at 17:56
Ceren BülbülCeren Bülbül
54
asked Mar 31 at 17:56
Ceren BülbülCeren Bülbül
54
54
closed as off-topic by Eevee Trainer, José Carlos Santos, John Omielan, Martin Argerami, Tianlalu Apr 1 at 4:12
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, José Carlos Santos, John Omielan, Martin Argerami, Tianlalu
closed as off-topic by Eevee Trainer, José Carlos Santos, John Omielan, Martin Argerami, Tianlalu Apr 1 at 4:12
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, José Carlos Santos, John Omielan, Martin Argerami, Tianlalu
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
To find the volume of the solid of revolution formed by rotating the bounded region $y=ax^2+1,y=0,x=0,x=1$ about the $x$-axis, use the disk method:
$$V_1(a)=int_a^bA(x)dx$$ where $A(x):=pi r^2$ is the cross-sectional area of a disk. Upon graphing the region, $$A(x)=pi(ax^2+1)^2=pi a^2x^4+2pi ax^2+pi$$Hence $$V_1(a)=piint_0^1left(a^2x^4+2ax^2+1right)dxspacetextfor $ageq-1$$$ $$=pileft(fraca^2x^55+frac2ax^33+xright)bigg
|^1_0=pileft(fraca^25+frac2a3+1right)$$
To find the volume of the solid of revolution formed by rotating the bounded region $y=ax^2+1,y=0,x=0,x=1$ about the $y$-axis, use the cylindrical shells method:
$$V_2(a)=int_a^b2pi xf(x)dx$$ Now we find the volume of a typical cylindrical shell: $$V=2pi rhDelta r=2pi(1)(f(x))dx$$ Note that the radius of the cylindrical shell is fixed at $1$ since the region is bound by $x=0$ and $x=1$.
Therefore $$V_2(a)=int_0^12pi(1)f(x) space dx=2pi int_0^1left(ax^2+1right)space dxspacetextfor $ageq-1$$$ $$=2pileft(fracax^33+xright)bigg|^1_0=pileft(frac2a3+2right)$$
For which $a$ is $V_1(a)=V_2(a)?impliespileft(fraca^25+frac2a3+1right)=pileft(frac2a3+2right).$ We have: $$pileft(fraca^25+frac2a3+1right)=pileft(frac2a3+2right)impliesfraca^25+frac2a3+1=frac2a3+2implies a^2=5implies$$ $$a=pmsqrt5.$$
Choose the positive root $a=sqrt5$ since $a=-sqrt5$ does not satisfy $ageq-1$.
answered Mar 31 at 18:24
coreyman317coreyman317
1,280522
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
To find the volume of the solid of revolution formed by rotating the bounded region $y=ax^2+1,y=0,x=0,x=1$ about the $x$-axis, use the disk method:
$$V_1(a)=int_a^bA(x)dx$$ where $A(x):=pi r^2$ is the cross-sectional area of a disk. Upon graphing the region, $$A(x)=pi(ax^2+1)^2=pi a^2x^4+2pi ax^2+pi$$Hence $$V_1(a)=piint_0^1left(a^2x^4+2ax^2+1right)dxspacetextfor $ageq-1$$$ $$=pileft(fraca^2x^55+frac2ax^33+xright)bigg
|^1_0=pileft(fraca^25+frac2a3+1right)$$
To find the volume of the solid of revolution formed by rotating the bounded region $y=ax^2+1,y=0,x=0,x=1$ about the $y$-axis, use the cylindrical shells method:
$$V_2(a)=int_a^b2pi xf(x)dx$$ Now we find the volume of a typical cylindrical shell: $$V=2pi rhDelta r=2pi(1)(f(x))dx$$ Note that the radius of the cylindrical shell is fixed at $1$ since the region is bound by $x=0$ and $x=1$.
Therefore $$V_2(a)=int_0^12pi(1)f(x) space dx=2pi int_0^1left(ax^2+1right)space dxspacetextfor $ageq-1$$$ $$=2pileft(fracax^33+xright)bigg|^1_0=pileft(frac2a3+2right)$$
For which $a$ is $V_1(a)=V_2(a)?impliespileft(fraca^25+frac2a3+1right)=pileft(frac2a3+2right).$ We have: $$pileft(fraca^25+frac2a3+1right)=pileft(frac2a3+2right)impliesfraca^25+frac2a3+1=frac2a3+2implies a^2=5implies$$ $$a=pmsqrt5.$$
Choose the positive root $a=sqrt5$ since $a=-sqrt5$ does not satisfy $ageq-1$.
answered Mar 31 at 18:24
coreyman317coreyman317
1,280522
$endgroup$
add a comment |
$begingroup$
To find the volume of the solid of revolution formed by rotating the bounded region $y=ax^2+1,y=0,x=0,x=1$ about the $x$-axis, use the disk method:
$$V_1(a)=int_a^bA(x)dx$$ where $A(x):=pi r^2$ is the cross-sectional area of a disk. Upon graphing the region, $$A(x)=pi(ax^2+1)^2=pi a^2x^4+2pi ax^2+pi$$Hence $$V_1(a)=piint_0^1left(a^2x^4+2ax^2+1right)dxspacetextfor $ageq-1$$$ $$=pileft(fraca^2x^55+frac2ax^33+xright)bigg
|^1_0=pileft(fraca^25+frac2a3+1right)$$
To find the volume of the solid of revolution formed by rotating the bounded region $y=ax^2+1,y=0,x=0,x=1$ about the $y$-axis, use the cylindrical shells method:
$$V_2(a)=int_a^b2pi xf(x)dx$$ Now we find the volume of a typical cylindrical shell: $$V=2pi rhDelta r=2pi(1)(f(x))dx$$ Note that the radius of the cylindrical shell is fixed at $1$ since the region is bound by $x=0$ and $x=1$.
Therefore $$V_2(a)=int_0^12pi(1)f(x) space dx=2pi int_0^1left(ax^2+1right)space dxspacetextfor $ageq-1$$$ $$=2pileft(fracax^33+xright)bigg|^1_0=pileft(frac2a3+2right)$$
For which $a$ is $V_1(a)=V_2(a)?impliespileft(fraca^25+frac2a3+1right)=pileft(frac2a3+2right).$ We have: $$pileft(fraca^25+frac2a3+1right)=pileft(frac2a3+2right)impliesfraca^25+frac2a3+1=frac2a3+2implies a^2=5implies$$ $$a=pmsqrt5.$$
Choose the positive root $a=sqrt5$ since $a=-sqrt5$ does not satisfy $ageq-1$.
answered Mar 31 at 18:24
coreyman317coreyman317
1,280522
$endgroup$
add a comment |
$begingroup$
To find the volume of the solid of revolution formed by rotating the bounded region $y=ax^2+1,y=0,x=0,x=1$ about the $x$-axis, use the disk method:
$$V_1(a)=int_a^bA(x)dx$$ where $A(x):=pi r^2$ is the cross-sectional area of a disk. Upon graphing the region, $$A(x)=pi(ax^2+1)^2=pi a^2x^4+2pi ax^2+pi$$Hence $$V_1(a)=piint_0^1left(a^2x^4+2ax^2+1right)dxspacetextfor $ageq-1$$$ $$=pileft(fraca^2x^55+frac2ax^33+xright)bigg
|^1_0=pileft(fraca^25+frac2a3+1right)$$
To find the volume of the solid of revolution formed by rotating the bounded region $y=ax^2+1,y=0,x=0,x=1$ about the $y$-axis, use the cylindrical shells method:
$$V_2(a)=int_a^b2pi xf(x)dx$$ Now we find the volume of a typical cylindrical shell: $$V=2pi rhDelta r=2pi(1)(f(x))dx$$ Note that the radius of the cylindrical shell is fixed at $1$ since the region is bound by $x=0$ and $x=1$.
Therefore $$V_2(a)=int_0^12pi(1)f(x) space dx=2pi int_0^1left(ax^2+1right)space dxspacetextfor $ageq-1$$$ $$=2pileft(fracax^33+xright)bigg|^1_0=pileft(frac2a3+2right)$$
For which $a$ is $V_1(a)=V_2(a)?impliespileft(fraca^25+frac2a3+1right)=pileft(frac2a3+2right).$ We have: $$pileft(fraca^25+frac2a3+1right)=pileft(frac2a3+2right)impliesfraca^25+frac2a3+1=frac2a3+2implies a^2=5implies$$ $$a=pmsqrt5.$$
Choose the positive root $a=sqrt5$ since $a=-sqrt5$ does not satisfy $ageq-1$.
answered Mar 31 at 18:24
coreyman317coreyman317
1,280522
$endgroup$
To find the volume of the solid of revolution formed by rotating the bounded region $y=ax^2+1,y=0,x=0,x=1$ about the $x$-axis, use the disk method:
$$V_1(a)=int_a^bA(x)dx$$ where $A(x):=pi r^2$ is the cross-sectional area of a disk. Upon graphing the region, $$A(x)=pi(ax^2+1)^2=pi a^2x^4+2pi ax^2+pi$$Hence $$V_1(a)=piint_0^1left(a^2x^4+2ax^2+1right)dxspacetextfor $ageq-1$$$ $$=pileft(fraca^2x^55+frac2ax^33+xright)bigg
|^1_0=pileft(fraca^25+frac2a3+1right)$$
To find the volume of the solid of revolution formed by rotating the bounded region $y=ax^2+1,y=0,x=0,x=1$ about the $y$-axis, use the cylindrical shells method:
$$V_2(a)=int_a^b2pi xf(x)dx$$ Now we find the volume of a typical cylindrical shell: $$V=2pi rhDelta r=2pi(1)(f(x))dx$$ Note that the radius of the cylindrical shell is fixed at $1$ since the region is bound by $x=0$ and $x=1$.
Therefore $$V_2(a)=int_0^12pi(1)f(x) space dx=2pi int_0^1left(ax^2+1right)space dxspacetextfor $ageq-1$$$ $$=2pileft(fracax^33+xright)bigg|^1_0=pileft(frac2a3+2right)$$
For which $a$ is $V_1(a)=V_2(a)?impliespileft(fraca^25+frac2a3+1right)=pileft(frac2a3+2right).$ We have: $$pileft(fraca^25+frac2a3+1right)=pileft(frac2a3+2right)impliesfraca^25+frac2a3+1=frac2a3+2implies a^2=5implies$$ $$a=pmsqrt5.$$
Choose the positive root $a=sqrt5$ since $a=-sqrt5$ does not satisfy $ageq-1$.
answered Mar 31 at 18:24
coreyman317coreyman317
1,280522
edited Mar 31 at 18:37
answered Mar 31 at 18:24
coreyman317coreyman317
1,280522
answered Mar 31 at 18:24
coreyman317coreyman317
1,280522
answered Mar 31 at 18:24
coreyman317coreyman317
1,280522
1,280522
add a comment |
add a comment |
