Volume of Revolution for $y=1-x^2$ and $y=2x$
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There's a question on our review sheet that I don't quite understand. It asks for the integral to find a volume of a solid bounded by the graphs $y=1-x^2$, $x=0$, and $y=2x$. The correct answer is $int_{0}^{sqrt2-1} pi[(1-x^2)^2-(2x)^2]dx$. But since they obviously had to put everything in terms of y to avoid doing two integrals and to get those bounds, shouldn't the answer have the inverse functions $int_{0}^{sqrt2-1} pi[(1-y)-(y/2)^2]dy$? Or am I missing something about the washer method?
calculus solid-of-revolution
asked Dec 8 '18 at 20:09
user3896106user3896106
83
$endgroup$
add a comment |
$begingroup$
There's a question on our review sheet that I don't quite understand. It asks for the integral to find a volume of a solid bounded by the graphs $y=1-x^2$, $x=0$, and $y=2x$. The correct answer is $int_{0}^{sqrt2-1} pi[(1-x^2)^2-(2x)^2]dx$. But since they obviously had to put everything in terms of y to avoid doing two integrals and to get those bounds, shouldn't the answer have the inverse functions $int_{0}^{sqrt2-1} pi[(1-y)-(y/2)^2]dy$? Or am I missing something about the washer method?
calculus solid-of-revolution
asked Dec 8 '18 at 20:09
user3896106user3896106
83
$endgroup$
add a comment |
$begingroup$
There's a question on our review sheet that I don't quite understand. It asks for the integral to find a volume of a solid bounded by the graphs $y=1-x^2$, $x=0$, and $y=2x$. The correct answer is $int_{0}^{sqrt2-1} pi[(1-x^2)^2-(2x)^2]dx$. But since they obviously had to put everything in terms of y to avoid doing two integrals and to get those bounds, shouldn't the answer have the inverse functions $int_{0}^{sqrt2-1} pi[(1-y)-(y/2)^2]dy$? Or am I missing something about the washer method?
calculus solid-of-revolution
asked Dec 8 '18 at 20:09
user3896106user3896106
83
$endgroup$
There's a question on our review sheet that I don't quite understand. It asks for the integral to find a volume of a solid bounded by the graphs $y=1-x^2$, $x=0$, and $y=2x$. The correct answer is $int_{0}^{sqrt2-1} pi[(1-x^2)^2-(2x)^2]dx$. But since they obviously had to put everything in terms of y to avoid doing two integrals and to get those bounds, shouldn't the answer have the inverse functions $int_{0}^{sqrt2-1} pi[(1-y)-(y/2)^2]dy$? Or am I missing something about the washer method?
calculus solid-of-revolution
calculus solid-of-revolution
asked Dec 8 '18 at 20:09
user3896106user3896106
83
asked Dec 8 '18 at 20:09
user3896106user3896106
83
edited Dec 8 '18 at 20:14
user3896106
asked Dec 8 '18 at 20:09
user3896106user3896106
83
asked Dec 8 '18 at 20:09
user3896106user3896106
83
asked Dec 8 '18 at 20:09
user3896106user3896106
83
83
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Try graphing the two functions. Inverses are unnecessary. This is a straight forward application of the disc (or as you put it "washer") method. This is because you can compute both volumes by the disc method and then subtract.
answered Dec 8 '18 at 20:16
Chris CusterChris Custer
12.4k3825
$endgroup$
$begingroup$
It seems I'm am either very tired or unable to read. I misinterpreted x=0 for y=0, so I though that it was asking for the solid of revolution of the between 0 and 1. That makes the problem much easier. Thanks for the help.
$endgroup$
– user3896106
Dec 8 '18 at 20:20
$begingroup$
No problem. You need of course to find the point of intersection of the two curves- that accounts for the limits.
$endgroup$
– Chris Custer
Dec 8 '18 at 20:43
add a comment |
$begingroup$
The integral
$$int_{0}^{sqrt2-1} pi[(1-x^2)^2-(2x)^2]dx$$
represents the volume for the revolution around the $x$ axis volume of the area bounded between the two functions for $xin [0,sqrt 2-1]$.
Refer also to the following sketch
answered Dec 8 '18 at 20:13
gimusigimusi
92.8k84494
$endgroup$
$begingroup$
How do you decide of the axis of rotation from the OP?
$endgroup$
– DJohnM
Dec 8 '18 at 21:57
$begingroup$
@DJohnM For the given set up otherwise the integral would be different and in 2 parts.
$endgroup$
– gimusi
Dec 8 '18 at 22:01
$begingroup$
So you use the given answer to decide what the missing information in the question actually is?
$endgroup$
– DJohnM
Dec 8 '18 at 22:17
$begingroup$
@DJohnM The OP is asking about: "There's a question on our review sheet that I don't quite understand...". Therefore I've explained what the given set up means and how it is obtained. Do not you agree with that interpretation?
$endgroup$
– gimusi
Dec 8 '18 at 22:22
$begingroup$
Yes, I agree...
$endgroup$
– DJohnM
Dec 8 '18 at 22:44
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Try graphing the two functions. Inverses are unnecessary. This is a straight forward application of the disc (or as you put it "washer") method. This is because you can compute both volumes by the disc method and then subtract.
answered Dec 8 '18 at 20:16
Chris CusterChris Custer
12.4k3825
$endgroup$
$begingroup$
It seems I'm am either very tired or unable to read. I misinterpreted x=0 for y=0, so I though that it was asking for the solid of revolution of the between 0 and 1. That makes the problem much easier. Thanks for the help.
$endgroup$
– user3896106
Dec 8 '18 at 20:20
$begingroup$
No problem. You need of course to find the point of intersection of the two curves- that accounts for the limits.
$endgroup$
– Chris Custer
Dec 8 '18 at 20:43
add a comment |
$begingroup$
Try graphing the two functions. Inverses are unnecessary. This is a straight forward application of the disc (or as you put it "washer") method. This is because you can compute both volumes by the disc method and then subtract.
answered Dec 8 '18 at 20:16
Chris CusterChris Custer
12.4k3825
$endgroup$
$begingroup$
It seems I'm am either very tired or unable to read. I misinterpreted x=0 for y=0, so I though that it was asking for the solid of revolution of the between 0 and 1. That makes the problem much easier. Thanks for the help.
$endgroup$
– user3896106
Dec 8 '18 at 20:20
$begingroup$
No problem. You need of course to find the point of intersection of the two curves- that accounts for the limits.
$endgroup$
– Chris Custer
Dec 8 '18 at 20:43
add a comment |
$begingroup$
Try graphing the two functions. Inverses are unnecessary. This is a straight forward application of the disc (or as you put it "washer") method. This is because you can compute both volumes by the disc method and then subtract.
answered Dec 8 '18 at 20:16
Chris CusterChris Custer
12.4k3825
$endgroup$
Try graphing the two functions. Inverses are unnecessary. This is a straight forward application of the disc (or as you put it "washer") method. This is because you can compute both volumes by the disc method and then subtract.
answered Dec 8 '18 at 20:16
Chris CusterChris Custer
12.4k3825
answered Dec 8 '18 at 20:16
Chris CusterChris Custer
12.4k3825
answered Dec 8 '18 at 20:16
Chris CusterChris Custer
12.4k3825
answered Dec 8 '18 at 20:16
Chris CusterChris Custer
12.4k3825
12.4k3825
$begingroup$
It seems I'm am either very tired or unable to read. I misinterpreted x=0 for y=0, so I though that it was asking for the solid of revolution of the between 0 and 1. That makes the problem much easier. Thanks for the help.
$endgroup$
– user3896106
Dec 8 '18 at 20:20
$begingroup$
No problem. You need of course to find the point of intersection of the two curves- that accounts for the limits.
$endgroup$
– Chris Custer
Dec 8 '18 at 20:43
add a comment |
$begingroup$
It seems I'm am either very tired or unable to read. I misinterpreted x=0 for y=0, so I though that it was asking for the solid of revolution of the between 0 and 1. That makes the problem much easier. Thanks for the help.
$endgroup$
– user3896106
Dec 8 '18 at 20:20
$begingroup$
No problem. You need of course to find the point of intersection of the two curves- that accounts for the limits.
$endgroup$
– Chris Custer
Dec 8 '18 at 20:43
$begingroup$
It seems I'm am either very tired or unable to read. I misinterpreted x=0 for y=0, so I though that it was asking for the solid of revolution of the between 0 and 1. That makes the problem much easier. Thanks for the help.
$endgroup$
– user3896106
Dec 8 '18 at 20:20
$begingroup$
It seems I'm am either very tired or unable to read. I misinterpreted x=0 for y=0, so I though that it was asking for the solid of revolution of the between 0 and 1. That makes the problem much easier. Thanks for the help.
$endgroup$
– user3896106
Dec 8 '18 at 20:20
$begingroup$
No problem. You need of course to find the point of intersection of the two curves- that accounts for the limits.
$endgroup$
– Chris Custer
Dec 8 '18 at 20:43
$begingroup$
No problem. You need of course to find the point of intersection of the two curves- that accounts for the limits.
$endgroup$
– Chris Custer
Dec 8 '18 at 20:43
add a comment |
$begingroup$
The integral
$$int_{0}^{sqrt2-1} pi[(1-x^2)^2-(2x)^2]dx$$
represents the volume for the revolution around the $x$ axis volume of the area bounded between the two functions for $xin [0,sqrt 2-1]$.
Refer also to the following sketch
answered Dec 8 '18 at 20:13
gimusigimusi
92.8k84494
$endgroup$
$begingroup$
How do you decide of the axis of rotation from the OP?
$endgroup$
– DJohnM
Dec 8 '18 at 21:57
$begingroup$
@DJohnM For the given set up otherwise the integral would be different and in 2 parts.
$endgroup$
– gimusi
Dec 8 '18 at 22:01
$begingroup$
So you use the given answer to decide what the missing information in the question actually is?
$endgroup$
– DJohnM
Dec 8 '18 at 22:17
$begingroup$
@DJohnM The OP is asking about: "There's a question on our review sheet that I don't quite understand...". Therefore I've explained what the given set up means and how it is obtained. Do not you agree with that interpretation?
$endgroup$
– gimusi
Dec 8 '18 at 22:22
$begingroup$
Yes, I agree...
$endgroup$
– DJohnM
Dec 8 '18 at 22:44
add a comment |
$begingroup$
The integral
$$int_{0}^{sqrt2-1} pi[(1-x^2)^2-(2x)^2]dx$$
represents the volume for the revolution around the $x$ axis volume of the area bounded between the two functions for $xin [0,sqrt 2-1]$.
Refer also to the following sketch
answered Dec 8 '18 at 20:13
gimusigimusi
92.8k84494
$endgroup$
$begingroup$
How do you decide of the axis of rotation from the OP?
$endgroup$
– DJohnM
Dec 8 '18 at 21:57
$begingroup$
@DJohnM For the given set up otherwise the integral would be different and in 2 parts.
$endgroup$
– gimusi
Dec 8 '18 at 22:01
$begingroup$
So you use the given answer to decide what the missing information in the question actually is?
$endgroup$
– DJohnM
Dec 8 '18 at 22:17
$begingroup$
@DJohnM The OP is asking about: "There's a question on our review sheet that I don't quite understand...". Therefore I've explained what the given set up means and how it is obtained. Do not you agree with that interpretation?
$endgroup$
– gimusi
Dec 8 '18 at 22:22
$begingroup$
Yes, I agree...
$endgroup$
– DJohnM
Dec 8 '18 at 22:44
add a comment |
$begingroup$
The integral
$$int_{0}^{sqrt2-1} pi[(1-x^2)^2-(2x)^2]dx$$
represents the volume for the revolution around the $x$ axis volume of the area bounded between the two functions for $xin [0,sqrt 2-1]$.
Refer also to the following sketch
answered Dec 8 '18 at 20:13
gimusigimusi
92.8k84494
$endgroup$
The integral
$$int_{0}^{sqrt2-1} pi[(1-x^2)^2-(2x)^2]dx$$
represents the volume for the revolution around the $x$ axis volume of the area bounded between the two functions for $xin [0,sqrt 2-1]$.
Refer also to the following sketch
answered Dec 8 '18 at 20:13
gimusigimusi
92.8k84494
edited Dec 8 '18 at 20:20
answered Dec 8 '18 at 20:13
gimusigimusi
92.8k84494
answered Dec 8 '18 at 20:13
gimusigimusi
92.8k84494
answered Dec 8 '18 at 20:13
gimusigimusi
92.8k84494
92.8k84494
$begingroup$
How do you decide of the axis of rotation from the OP?
$endgroup$
– DJohnM
Dec 8 '18 at 21:57
$begingroup$
@DJohnM For the given set up otherwise the integral would be different and in 2 parts.
$endgroup$
– gimusi
Dec 8 '18 at 22:01
$begingroup$
So you use the given answer to decide what the missing information in the question actually is?
$endgroup$
– DJohnM
Dec 8 '18 at 22:17
$begingroup$
@DJohnM The OP is asking about: "There's a question on our review sheet that I don't quite understand...". Therefore I've explained what the given set up means and how it is obtained. Do not you agree with that interpretation?
$endgroup$
– gimusi
Dec 8 '18 at 22:22
$begingroup$
Yes, I agree...
$endgroup$
– DJohnM
Dec 8 '18 at 22:44
add a comment |
$begingroup$
How do you decide of the axis of rotation from the OP?
$endgroup$
– DJohnM
Dec 8 '18 at 21:57
$begingroup$
@DJohnM For the given set up otherwise the integral would be different and in 2 parts.
$endgroup$
– gimusi
Dec 8 '18 at 22:01
$begingroup$
So you use the given answer to decide what the missing information in the question actually is?
$endgroup$
– DJohnM
Dec 8 '18 at 22:17
$begingroup$
@DJohnM The OP is asking about: "There's a question on our review sheet that I don't quite understand...". Therefore I've explained what the given set up means and how it is obtained. Do not you agree with that interpretation?
$endgroup$
– gimusi
Dec 8 '18 at 22:22
$begingroup$
Yes, I agree...
$endgroup$
– DJohnM
Dec 8 '18 at 22:44
$begingroup$
How do you decide of the axis of rotation from the OP?
$endgroup$
– DJohnM
Dec 8 '18 at 21:57
$begingroup$
How do you decide of the axis of rotation from the OP?
$endgroup$
– DJohnM
Dec 8 '18 at 21:57
$begingroup$
@DJohnM For the given set up otherwise the integral would be different and in 2 parts.
$endgroup$
– gimusi
Dec 8 '18 at 22:01
$begingroup$
@DJohnM For the given set up otherwise the integral would be different and in 2 parts.
$endgroup$
– gimusi
Dec 8 '18 at 22:01
$begingroup$
So you use the given answer to decide what the missing information in the question actually is?
$endgroup$
– DJohnM
Dec 8 '18 at 22:17
$begingroup$
So you use the given answer to decide what the missing information in the question actually is?
$endgroup$
– DJohnM
Dec 8 '18 at 22:17
$begingroup$
@DJohnM The OP is asking about: "There's a question on our review sheet that I don't quite understand...". Therefore I've explained what the given set up means and how it is obtained. Do not you agree with that interpretation?
$endgroup$
– gimusi
Dec 8 '18 at 22:22
$begingroup$
@DJohnM The OP is asking about: "There's a question on our review sheet that I don't quite understand...". Therefore I've explained what the given set up means and how it is obtained. Do not you agree with that interpretation?
$endgroup$
– gimusi
Dec 8 '18 at 22:22
$begingroup$
Yes, I agree...
$endgroup$
– DJohnM
Dec 8 '18 at 22:44
$begingroup$
Yes, I agree...
$endgroup$
– DJohnM
Dec 8 '18 at 22:44
add a comment |
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