Integral Calculus. Overlapping volume The Next CEO of Stack OverflowVolume of a rotated region?Volume of revolution: integral calculusVolume problem: Is my integral correct?Volume of a solid.Find the volume of the solid formed by the revolving the region around a lineVolume of the solid generated by revolving the region R enclosed by the curve - Disk and Shell methodFind volume of a solidCalculus volume questionFind the volume of the solid formed when a region is rotated about the $y$-axisUsing the washer method find the volume of the solid generated by the enclosed region
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Integral Calculus. Overlapping volume
The Next CEO of Stack OverflowVolume of a rotated region?Volume of revolution: integral calculusVolume problem: Is my integral correct?Volume of a solid.Find the volume of the solid formed by the revolving the region around a lineVolume of the solid generated by revolving the region R enclosed by the curve - Disk and Shell methodFind volume of a solidCalculus volume questionFind the volume of the solid formed when a region is rotated about the $y$-axisUsing the washer method find the volume of the solid generated by the enclosed region
$begingroup$
I wondering whether there is a more efficient and not overly complicated way to approach the solution...
If the volume V of the solid generated when the region enclosed by the curve
$y=x^2-2x$ and the line $y=2x$ is rotated about the x-axis.
[Sol] Finding the x-coordinates of the point of intersection of the curve and the line
$x^2-2x=2x$
$x^2-4x=0$
$x(x-4)=0$
Therefore, $x=0,4$
When the region is rotated about the x-axis, there is an overlapping part. This overlapping is caused by the region below the x-axis and should be disregarded in the calculation of the volume
Volume = $V=pi int^4_0[2x]^2 dx - pi int^4_2[x^2-2x]^2 dx$
=$pi int^4_0[4x] dx - pi int^4_2[x^4-4x^3+4x^2] dx$
=$pi left.left[frac4x^33 right] right|^4_0$ - $pi left.left[fracx^55-x^4+frac4x^33 right] right|^4_2$
=$pi(frac2563-0)-pi(frac10245-256+frac2563)-(frac325-16+frac323)]$
=$frac78415 pi$
calculus integration
edited Mar 18 at 11:03
YuiTo Cheng
2,1412937
asked Mar 18 at 10:36
R.SuR.Su
112
$endgroup$
add a comment |
$begingroup$
I wondering whether there is a more efficient and not overly complicated way to approach the solution...
If the volume V of the solid generated when the region enclosed by the curve
$y=x^2-2x$ and the line $y=2x$ is rotated about the x-axis.
[Sol] Finding the x-coordinates of the point of intersection of the curve and the line
$x^2-2x=2x$
$x^2-4x=0$
$x(x-4)=0$
Therefore, $x=0,4$
When the region is rotated about the x-axis, there is an overlapping part. This overlapping is caused by the region below the x-axis and should be disregarded in the calculation of the volume
Volume = $V=pi int^4_0[2x]^2 dx - pi int^4_2[x^2-2x]^2 dx$
=$pi int^4_0[4x] dx - pi int^4_2[x^4-4x^3+4x^2] dx$
=$pi left.left[frac4x^33 right] right|^4_0$ - $pi left.left[fracx^55-x^4+frac4x^33 right] right|^4_2$
=$pi(frac2563-0)-pi(frac10245-256+frac2563)-(frac325-16+frac323)]$
=$frac78415 pi$
calculus integration
edited Mar 18 at 11:03
YuiTo Cheng
2,1412937
asked Mar 18 at 10:36
R.SuR.Su
112
$endgroup$
$begingroup$
Please use MathJax, this is unreadable.
$endgroup$
– Klaus
Mar 18 at 10:37
1
$begingroup$
Sorry about that. I've still new to Mathjax. Hope this is slightly better
$endgroup$
– R.Su
Mar 18 at 10:48
add a comment |
$begingroup$
I wondering whether there is a more efficient and not overly complicated way to approach the solution...
If the volume V of the solid generated when the region enclosed by the curve
$y=x^2-2x$ and the line $y=2x$ is rotated about the x-axis.
[Sol] Finding the x-coordinates of the point of intersection of the curve and the line
$x^2-2x=2x$
$x^2-4x=0$
$x(x-4)=0$
Therefore, $x=0,4$
When the region is rotated about the x-axis, there is an overlapping part. This overlapping is caused by the region below the x-axis and should be disregarded in the calculation of the volume
Volume = $V=pi int^4_0[2x]^2 dx - pi int^4_2[x^2-2x]^2 dx$
=$pi int^4_0[4x] dx - pi int^4_2[x^4-4x^3+4x^2] dx$
=$pi left.left[frac4x^33 right] right|^4_0$ - $pi left.left[fracx^55-x^4+frac4x^33 right] right|^4_2$
=$pi(frac2563-0)-pi(frac10245-256+frac2563)-(frac325-16+frac323)]$
=$frac78415 pi$
calculus integration
edited Mar 18 at 11:03
YuiTo Cheng
2,1412937
asked Mar 18 at 10:36
R.SuR.Su
112
$endgroup$
I wondering whether there is a more efficient and not overly complicated way to approach the solution...
If the volume V of the solid generated when the region enclosed by the curve
$y=x^2-2x$ and the line $y=2x$ is rotated about the x-axis.
[Sol] Finding the x-coordinates of the point of intersection of the curve and the line
$x^2-2x=2x$
$x^2-4x=0$
$x(x-4)=0$
Therefore, $x=0,4$
When the region is rotated about the x-axis, there is an overlapping part. This overlapping is caused by the region below the x-axis and should be disregarded in the calculation of the volume
Volume = $V=pi int^4_0[2x]^2 dx - pi int^4_2[x^2-2x]^2 dx$
=$pi int^4_0[4x] dx - pi int^4_2[x^4-4x^3+4x^2] dx$
=$pi left.left[frac4x^33 right] right|^4_0$ - $pi left.left[fracx^55-x^4+frac4x^33 right] right|^4_2$
=$pi(frac2563-0)-pi(frac10245-256+frac2563)-(frac325-16+frac323)]$
=$frac78415 pi$
calculus integration
calculus integration
edited Mar 18 at 11:03
YuiTo Cheng
2,1412937
asked Mar 18 at 10:36
R.SuR.Su
112
edited Mar 18 at 11:03
YuiTo Cheng
2,1412937
asked Mar 18 at 10:36
R.SuR.Su
112
edited Mar 18 at 11:03
YuiTo Cheng
2,1412937
edited Mar 18 at 11:03
YuiTo Cheng
2,1412937
edited Mar 18 at 11:03
YuiTo Cheng
2,1412937
2,1412937
asked Mar 18 at 10:36
R.SuR.Su
112
asked Mar 18 at 10:36
R.SuR.Su
112
asked Mar 18 at 10:36
R.SuR.Su
112
112
$begingroup$
Please use MathJax, this is unreadable.
$endgroup$
– Klaus
Mar 18 at 10:37
1
$begingroup$
Sorry about that. I've still new to Mathjax. Hope this is slightly better
$endgroup$
– R.Su
Mar 18 at 10:48
add a comment |
$begingroup$
Please use MathJax, this is unreadable.
$endgroup$
– Klaus
Mar 18 at 10:37
1
$begingroup$
Sorry about that. I've still new to Mathjax. Hope this is slightly better
$endgroup$
– R.Su
Mar 18 at 10:48
$begingroup$
Please use MathJax, this is unreadable.
$endgroup$
– Klaus
Mar 18 at 10:37
$begingroup$
Please use MathJax, this is unreadable.
$endgroup$
– Klaus
Mar 18 at 10:37
1
1
$begingroup$
Sorry about that. I've still new to Mathjax. Hope this is slightly better
$endgroup$
– R.Su
Mar 18 at 10:48
$begingroup$
Sorry about that. I've still new to Mathjax. Hope this is slightly better
$endgroup$
– R.Su
Mar 18 at 10:48
add a comment |
0
active
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$begingroup$
Please use MathJax, this is unreadable.
$endgroup$
– Klaus
Mar 18 at 10:37
1
$begingroup$
Sorry about that. I've still new to Mathjax. Hope this is slightly better
$endgroup$
– R.Su
Mar 18 at 10:48