Using the washer method find the volume of the solid generated by the enclosed region The 2019...
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Using the washer method find the volume of the solid generated by the enclosed region
The 2019 Stack Overflow Developer Survey Results Are InFind the volume of the solid obtained by rotating the region enclosed by the curves $y=x^2 , x = 1, x = 2$, and $y=0$ about the line $x=5$Finding Volume of the Solid--washer methodFind Volume of a solid when the region is revolved about y-axis or x-axisRegion bounded by the parabola $, y=x^2;$ and the line $, y=16;$ What is the volume of the solid generated when R is rotated about $y = 17$?Revolving about the y-axisUse the washer method to find the volume of an area rotated about y=4Volume of the solid generated by revolving the region R enclosed by the curve - Disk and Shell methodFind volume of a solidFind the volume of the resulting solid (understanding to find radius)Solid of Revolution using Disk or Washer Method
$begingroup$
Find the volume of the solid generated when the region enclosed by $y=e^x$ and $y=ln x$ between $x=1/2$ and $x=1$ is revolved about the line $x=-4$.
I've tried setting it up using the standard of the outer radius subtracted by the inner radius but I'm not sure if it's set right because the region is enclosed by intersections.
$$pi int _{frac{1}{2}}^1:left(e^x-left(-4right)right)^2-left(lnleft(xright)-left(-4right)right)^2,dx$$
calculus
edited Feb 11 at 16:57
Exp ikx
4389
asked Feb 11 at 16:10
Tylor GonzalezTylor Gonzalez
12
$endgroup$
add a comment |
$begingroup$
Find the volume of the solid generated when the region enclosed by $y=e^x$ and $y=ln x$ between $x=1/2$ and $x=1$ is revolved about the line $x=-4$.
I've tried setting it up using the standard of the outer radius subtracted by the inner radius but I'm not sure if it's set right because the region is enclosed by intersections.
$$pi int _{frac{1}{2}}^1:left(e^x-left(-4right)right)^2-left(lnleft(xright)-left(-4right)right)^2,dx$$
calculus
edited Feb 11 at 16:57
Exp ikx
4389
asked Feb 11 at 16:10
Tylor GonzalezTylor Gonzalez
12
$endgroup$
add a comment |
$begingroup$
Find the volume of the solid generated when the region enclosed by $y=e^x$ and $y=ln x$ between $x=1/2$ and $x=1$ is revolved about the line $x=-4$.
I've tried setting it up using the standard of the outer radius subtracted by the inner radius but I'm not sure if it's set right because the region is enclosed by intersections.
$$pi int _{frac{1}{2}}^1:left(e^x-left(-4right)right)^2-left(lnleft(xright)-left(-4right)right)^2,dx$$
calculus
edited Feb 11 at 16:57
Exp ikx
4389
asked Feb 11 at 16:10
Tylor GonzalezTylor Gonzalez
12
$endgroup$
Find the volume of the solid generated when the region enclosed by $y=e^x$ and $y=ln x$ between $x=1/2$ and $x=1$ is revolved about the line $x=-4$.
I've tried setting it up using the standard of the outer radius subtracted by the inner radius but I'm not sure if it's set right because the region is enclosed by intersections.
$$pi int _{frac{1}{2}}^1:left(e^x-left(-4right)right)^2-left(lnleft(xright)-left(-4right)right)^2,dx$$
calculus
calculus
edited Feb 11 at 16:57
Exp ikx
4389
asked Feb 11 at 16:10
Tylor GonzalezTylor Gonzalez
12
edited Feb 11 at 16:57
Exp ikx
4389
asked Feb 11 at 16:10
Tylor GonzalezTylor Gonzalez
12
edited Feb 11 at 16:57
Exp ikx
4389
edited Feb 11 at 16:57
Exp ikx
4389
edited Feb 11 at 16:57
Exp ikx
4389
4389
asked Feb 11 at 16:10
Tylor GonzalezTylor Gonzalez
12
asked Feb 11 at 16:10
Tylor GonzalezTylor Gonzalez
12
asked Feb 11 at 16:10
Tylor GonzalezTylor Gonzalez
12
12
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The washer method for revolving about the y axis is:
Consider x inside the region 1/2 and 1 and a infinitesimal width dx the washer dA is (neglecting $(dx)^2$)
$$dA = pi((4 + x + dx)^2 - (4 + x)^2)dx = 2pi (4 + x)dx$$
Hence the volume is
$$V = 2piint_{frac{1}{2}}^{1}(e^x - ln x)(4 + x)dx$$
$$ = 2piint_{frac{1}{2}}^{1}(4e^x + xe^x - 4ln x - xln x)dx$$
$$= 2pi(4e - frac{7}{2}sqrt{e} + frac{35}{16} - frac{17}{8}ln 2)$$
$$= 36.55$$
answered Mar 21 at 15:33
KY TangKY Tang
50436
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The washer method for revolving about the y axis is:
Consider x inside the region 1/2 and 1 and a infinitesimal width dx the washer dA is (neglecting $(dx)^2$)
$$dA = pi((4 + x + dx)^2 - (4 + x)^2)dx = 2pi (4 + x)dx$$
Hence the volume is
$$V = 2piint_{frac{1}{2}}^{1}(e^x - ln x)(4 + x)dx$$
$$ = 2piint_{frac{1}{2}}^{1}(4e^x + xe^x - 4ln x - xln x)dx$$
$$= 2pi(4e - frac{7}{2}sqrt{e} + frac{35}{16} - frac{17}{8}ln 2)$$
$$= 36.55$$
answered Mar 21 at 15:33
KY TangKY Tang
50436
$endgroup$
add a comment |
$begingroup$
The washer method for revolving about the y axis is:
Consider x inside the region 1/2 and 1 and a infinitesimal width dx the washer dA is (neglecting $(dx)^2$)
$$dA = pi((4 + x + dx)^2 - (4 + x)^2)dx = 2pi (4 + x)dx$$
Hence the volume is
$$V = 2piint_{frac{1}{2}}^{1}(e^x - ln x)(4 + x)dx$$
$$ = 2piint_{frac{1}{2}}^{1}(4e^x + xe^x - 4ln x - xln x)dx$$
$$= 2pi(4e - frac{7}{2}sqrt{e} + frac{35}{16} - frac{17}{8}ln 2)$$
$$= 36.55$$
answered Mar 21 at 15:33
KY TangKY Tang
50436
$endgroup$
add a comment |
$begingroup$
The washer method for revolving about the y axis is:
Consider x inside the region 1/2 and 1 and a infinitesimal width dx the washer dA is (neglecting $(dx)^2$)
$$dA = pi((4 + x + dx)^2 - (4 + x)^2)dx = 2pi (4 + x)dx$$
Hence the volume is
$$V = 2piint_{frac{1}{2}}^{1}(e^x - ln x)(4 + x)dx$$
$$ = 2piint_{frac{1}{2}}^{1}(4e^x + xe^x - 4ln x - xln x)dx$$
$$= 2pi(4e - frac{7}{2}sqrt{e} + frac{35}{16} - frac{17}{8}ln 2)$$
$$= 36.55$$
answered Mar 21 at 15:33
KY TangKY Tang
50436
$endgroup$
The washer method for revolving about the y axis is:
Consider x inside the region 1/2 and 1 and a infinitesimal width dx the washer dA is (neglecting $(dx)^2$)
$$dA = pi((4 + x + dx)^2 - (4 + x)^2)dx = 2pi (4 + x)dx$$
Hence the volume is
$$V = 2piint_{frac{1}{2}}^{1}(e^x - ln x)(4 + x)dx$$
$$ = 2piint_{frac{1}{2}}^{1}(4e^x + xe^x - 4ln x - xln x)dx$$
$$= 2pi(4e - frac{7}{2}sqrt{e} + frac{35}{16} - frac{17}{8}ln 2)$$
$$= 36.55$$
answered Mar 21 at 15:33
KY TangKY Tang
50436
answered Mar 21 at 15:33
KY TangKY Tang
50436
answered Mar 21 at 15:33
KY TangKY Tang
50436
answered Mar 21 at 15:33
KY TangKY Tang
50436
50436
add a comment |
add a comment |
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