Find the volume of the solid formed by rotating about the $x$-axis
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Find the volume of the solid formed by rotating the region enclosed by the curves $y=(e^ x) + 2$, $y=0$ , $x=0$, and $x=0.7$ about the $x$-axis
I set up the equations as follows using the washer method. I'm not sure if I'm setting it up right or using the correct method.
$$int_0^0.7 pi (e^x+2)^2dx$$
calculus volume
edited Feb 3 '14 at 19:37
AlexR
22.4k12348
asked Feb 3 '14 at 19:35
janny
901312
add a comment |Â
up vote
4
down vote
favorite
Find the volume of the solid formed by rotating the region enclosed by the curves $y=(e^ x) + 2$, $y=0$ , $x=0$, and $x=0.7$ about the $x$-axis
I set up the equations as follows using the washer method. I'm not sure if I'm setting it up right or using the correct method.
$$int_0^0.7 pi (e^x+2)^2dx$$
calculus volume
edited Feb 3 '14 at 19:37
AlexR
22.4k12348
asked Feb 3 '14 at 19:35
janny
901312
1
The setup is correct. To finish, expand $(e^x+2)^2$ and integrate.
– André Nicolas
Feb 3 '14 at 19:37
1
Note you can pull the $pi$ out of the integral.
– AlexR
Feb 3 '14 at 19:37
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Find the volume of the solid formed by rotating the region enclosed by the curves $y=(e^ x) + 2$, $y=0$ , $x=0$, and $x=0.7$ about the $x$-axis
I set up the equations as follows using the washer method. I'm not sure if I'm setting it up right or using the correct method.
$$int_0^0.7 pi (e^x+2)^2dx$$
calculus volume
edited Feb 3 '14 at 19:37
AlexR
22.4k12348
asked Feb 3 '14 at 19:35
janny
901312
Find the volume of the solid formed by rotating the region enclosed by the curves $y=(e^ x) + 2$, $y=0$ , $x=0$, and $x=0.7$ about the $x$-axis
I set up the equations as follows using the washer method. I'm not sure if I'm setting it up right or using the correct method.
$$int_0^0.7 pi (e^x+2)^2dx$$
calculus volume
edited Feb 3 '14 at 19:37
AlexR
22.4k12348
asked Feb 3 '14 at 19:35
janny
901312
edited Feb 3 '14 at 19:37
AlexR
22.4k12348
edited Feb 3 '14 at 19:37
AlexR
22.4k12348
edited Feb 3 '14 at 19:37
AlexR
22.4k12348
22.4k12348
asked Feb 3 '14 at 19:35
janny
901312
asked Feb 3 '14 at 19:35
janny
901312
asked Feb 3 '14 at 19:35
janny
901312
901312
1
The setup is correct. To finish, expand $(e^x+2)^2$ and integrate.
– André Nicolas
Feb 3 '14 at 19:37
1
Note you can pull the $pi$ out of the integral.
– AlexR
Feb 3 '14 at 19:37
add a comment |Â
1
The setup is correct. To finish, expand $(e^x+2)^2$ and integrate.
– André Nicolas
Feb 3 '14 at 19:37
1
Note you can pull the $pi$ out of the integral.
– AlexR
Feb 3 '14 at 19:37
1
1
The setup is correct. To finish, expand $(e^x+2)^2$ and integrate.
– André Nicolas
Feb 3 '14 at 19:37
The setup is correct. To finish, expand $(e^x+2)^2$ and integrate.
– André Nicolas
Feb 3 '14 at 19:37
1
1
Note you can pull the $pi$ out of the integral.
– AlexR
Feb 3 '14 at 19:37
Note you can pull the $pi$ out of the integral.
– AlexR
Feb 3 '14 at 19:37
add a comment |Â
2 Answers
2
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oldest
votes
up vote
0
down vote
Your equation is correct.
$$beginalign*
A & = pi int_0^0.7 (e^x + 2)^2 dx = pi left( int_0^0.7 e^2x dx + 4 int_0^0.7 e^x dx + 4cdot0.7 right)\
& = pi left( frac12 int_0^1.4 e^x dx + int_0^0.7 e^x dx + 2.8right)\
& = pi left( frac12 e^1.4 + e^0.7 + 2.8 right) \
& approx 21.4927ldots
endalign*$$
answered Feb 3 '14 at 19:41
AlexR
22.4k12348
Not sure your answer is correct. Note $e^0ne0$.
– John Molokach
Jan 9 '16 at 14:59
@JohnMolokach $e^0 = 1$ by definition...
– AlexR
Jan 9 '16 at 15:00
Sorry, edited...
– John Molokach
Jan 9 '16 at 15:01
It might be me, but are you sure you can make the simplification of taking $e^2x$ out to the range over which you're integrating? As if you leave it in and integrate, it integrates to $e^2x^2$ so you wouldn't simply be evaluating $e^x$ between $0$ and $1.4$. I'm not entirely sure that gives the same answer.
– Robert Frost
May 17 at 10:46
@RobertFrost That‘s a substitution, yielding the factor of $1/2$
– AlexR
May 17 at 11:18
 |Â
show 2 more comments
up vote
0
down vote
Your setting up was all correct. However you can conveniently integrate with symbols and get on with integration to go to numerical work in the last stages of calculation by plugging in constants. Also bring out constants to LHS at the earliest, else that keeps dragging on till the end.
$$beginalign*
A/pi & = int_0^a (e^x + c)^2 dx = int_0^a e^2x dx + 2c int_0^a e^x dx + c^2 a\
& = frac12 ( e^2a-1) + 2c(e^a-1) dx + ac^2.\
endalign*$$
answered Apr 25 '17 at 9:39
Narasimham
20.2k51957
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Your equation is correct.
$$beginalign*
A & = pi int_0^0.7 (e^x + 2)^2 dx = pi left( int_0^0.7 e^2x dx + 4 int_0^0.7 e^x dx + 4cdot0.7 right)\
& = pi left( frac12 int_0^1.4 e^x dx + int_0^0.7 e^x dx + 2.8right)\
& = pi left( frac12 e^1.4 + e^0.7 + 2.8 right) \
& approx 21.4927ldots
endalign*$$
answered Feb 3 '14 at 19:41
AlexR
22.4k12348
Not sure your answer is correct. Note $e^0ne0$.
– John Molokach
Jan 9 '16 at 14:59
@JohnMolokach $e^0 = 1$ by definition...
– AlexR
Jan 9 '16 at 15:00
Sorry, edited...
– John Molokach
Jan 9 '16 at 15:01
It might be me, but are you sure you can make the simplification of taking $e^2x$ out to the range over which you're integrating? As if you leave it in and integrate, it integrates to $e^2x^2$ so you wouldn't simply be evaluating $e^x$ between $0$ and $1.4$. I'm not entirely sure that gives the same answer.
– Robert Frost
May 17 at 10:46
@RobertFrost That‘s a substitution, yielding the factor of $1/2$
– AlexR
May 17 at 11:18
 |Â
show 2 more comments
up vote
0
down vote
Your equation is correct.
$$beginalign*
A & = pi int_0^0.7 (e^x + 2)^2 dx = pi left( int_0^0.7 e^2x dx + 4 int_0^0.7 e^x dx + 4cdot0.7 right)\
& = pi left( frac12 int_0^1.4 e^x dx + int_0^0.7 e^x dx + 2.8right)\
& = pi left( frac12 e^1.4 + e^0.7 + 2.8 right) \
& approx 21.4927ldots
endalign*$$
answered Feb 3 '14 at 19:41
AlexR
22.4k12348
Not sure your answer is correct. Note $e^0ne0$.
– John Molokach
Jan 9 '16 at 14:59
@JohnMolokach $e^0 = 1$ by definition...
– AlexR
Jan 9 '16 at 15:00
Sorry, edited...
– John Molokach
Jan 9 '16 at 15:01
It might be me, but are you sure you can make the simplification of taking $e^2x$ out to the range over which you're integrating? As if you leave it in and integrate, it integrates to $e^2x^2$ so you wouldn't simply be evaluating $e^x$ between $0$ and $1.4$. I'm not entirely sure that gives the same answer.
– Robert Frost
May 17 at 10:46
@RobertFrost That‘s a substitution, yielding the factor of $1/2$
– AlexR
May 17 at 11:18
 |Â
show 2 more comments
up vote
0
down vote
up vote
0
down vote
Your equation is correct.
$$beginalign*
A & = pi int_0^0.7 (e^x + 2)^2 dx = pi left( int_0^0.7 e^2x dx + 4 int_0^0.7 e^x dx + 4cdot0.7 right)\
& = pi left( frac12 int_0^1.4 e^x dx + int_0^0.7 e^x dx + 2.8right)\
& = pi left( frac12 e^1.4 + e^0.7 + 2.8 right) \
& approx 21.4927ldots
endalign*$$
answered Feb 3 '14 at 19:41
AlexR
22.4k12348
Your equation is correct.
$$beginalign*
A & = pi int_0^0.7 (e^x + 2)^2 dx = pi left( int_0^0.7 e^2x dx + 4 int_0^0.7 e^x dx + 4cdot0.7 right)\
& = pi left( frac12 int_0^1.4 e^x dx + int_0^0.7 e^x dx + 2.8right)\
& = pi left( frac12 e^1.4 + e^0.7 + 2.8 right) \
& approx 21.4927ldots
endalign*$$
answered Feb 3 '14 at 19:41
AlexR
22.4k12348
answered Feb 3 '14 at 19:41
AlexR
22.4k12348
answered Feb 3 '14 at 19:41
AlexR
22.4k12348
answered Feb 3 '14 at 19:41
AlexR
22.4k12348
22.4k12348
Not sure your answer is correct. Note $e^0ne0$.
– John Molokach
Jan 9 '16 at 14:59
@JohnMolokach $e^0 = 1$ by definition...
– AlexR
Jan 9 '16 at 15:00
Sorry, edited...
– John Molokach
Jan 9 '16 at 15:01
It might be me, but are you sure you can make the simplification of taking $e^2x$ out to the range over which you're integrating? As if you leave it in and integrate, it integrates to $e^2x^2$ so you wouldn't simply be evaluating $e^x$ between $0$ and $1.4$. I'm not entirely sure that gives the same answer.
– Robert Frost
May 17 at 10:46
@RobertFrost That‘s a substitution, yielding the factor of $1/2$
– AlexR
May 17 at 11:18
 |Â
show 2 more comments
Not sure your answer is correct. Note $e^0ne0$.
– John Molokach
Jan 9 '16 at 14:59
@JohnMolokach $e^0 = 1$ by definition...
– AlexR
Jan 9 '16 at 15:00
Sorry, edited...
– John Molokach
Jan 9 '16 at 15:01
It might be me, but are you sure you can make the simplification of taking $e^2x$ out to the range over which you're integrating? As if you leave it in and integrate, it integrates to $e^2x^2$ so you wouldn't simply be evaluating $e^x$ between $0$ and $1.4$. I'm not entirely sure that gives the same answer.
– Robert Frost
May 17 at 10:46
@RobertFrost That‘s a substitution, yielding the factor of $1/2$
– AlexR
May 17 at 11:18
Not sure your answer is correct. Note $e^0ne0$.
– John Molokach
Jan 9 '16 at 14:59
Not sure your answer is correct. Note $e^0ne0$.
– John Molokach
Jan 9 '16 at 14:59
@JohnMolokach $e^0 = 1$ by definition...
– AlexR
Jan 9 '16 at 15:00
@JohnMolokach $e^0 = 1$ by definition...
– AlexR
Jan 9 '16 at 15:00
Sorry, edited...
– John Molokach
Jan 9 '16 at 15:01
Sorry, edited...
– John Molokach
Jan 9 '16 at 15:01
It might be me, but are you sure you can make the simplification of taking $e^2x$ out to the range over which you're integrating? As if you leave it in and integrate, it integrates to $e^2x^2$ so you wouldn't simply be evaluating $e^x$ between $0$ and $1.4$. I'm not entirely sure that gives the same answer.
– Robert Frost
May 17 at 10:46
It might be me, but are you sure you can make the simplification of taking $e^2x$ out to the range over which you're integrating? As if you leave it in and integrate, it integrates to $e^2x^2$ so you wouldn't simply be evaluating $e^x$ between $0$ and $1.4$. I'm not entirely sure that gives the same answer.
– Robert Frost
May 17 at 10:46
@RobertFrost That‘s a substitution, yielding the factor of $1/2$
– AlexR
May 17 at 11:18
@RobertFrost That‘s a substitution, yielding the factor of $1/2$
– AlexR
May 17 at 11:18
 |Â
show 2 more comments
up vote
0
down vote
Your setting up was all correct. However you can conveniently integrate with symbols and get on with integration to go to numerical work in the last stages of calculation by plugging in constants. Also bring out constants to LHS at the earliest, else that keeps dragging on till the end.
$$beginalign*
A/pi & = int_0^a (e^x + c)^2 dx = int_0^a e^2x dx + 2c int_0^a e^x dx + c^2 a\
& = frac12 ( e^2a-1) + 2c(e^a-1) dx + ac^2.\
endalign*$$
answered Apr 25 '17 at 9:39
Narasimham
20.2k51957
add a comment |Â
up vote
0
down vote
Your setting up was all correct. However you can conveniently integrate with symbols and get on with integration to go to numerical work in the last stages of calculation by plugging in constants. Also bring out constants to LHS at the earliest, else that keeps dragging on till the end.
$$beginalign*
A/pi & = int_0^a (e^x + c)^2 dx = int_0^a e^2x dx + 2c int_0^a e^x dx + c^2 a\
& = frac12 ( e^2a-1) + 2c(e^a-1) dx + ac^2.\
endalign*$$
answered Apr 25 '17 at 9:39
Narasimham
20.2k51957
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Your setting up was all correct. However you can conveniently integrate with symbols and get on with integration to go to numerical work in the last stages of calculation by plugging in constants. Also bring out constants to LHS at the earliest, else that keeps dragging on till the end.
$$beginalign*
A/pi & = int_0^a (e^x + c)^2 dx = int_0^a e^2x dx + 2c int_0^a e^x dx + c^2 a\
& = frac12 ( e^2a-1) + 2c(e^a-1) dx + ac^2.\
endalign*$$
answered Apr 25 '17 at 9:39
Narasimham
20.2k51957
Your setting up was all correct. However you can conveniently integrate with symbols and get on with integration to go to numerical work in the last stages of calculation by plugging in constants. Also bring out constants to LHS at the earliest, else that keeps dragging on till the end.
$$beginalign*
A/pi & = int_0^a (e^x + c)^2 dx = int_0^a e^2x dx + 2c int_0^a e^x dx + c^2 a\
& = frac12 ( e^2a-1) + 2c(e^a-1) dx + ac^2.\
endalign*$$
answered Apr 25 '17 at 9:39
Narasimham
20.2k51957
edited Apr 25 '17 at 11:23
answered Apr 25 '17 at 9:39
Narasimham
20.2k51957
answered Apr 25 '17 at 9:39
Narasimham
20.2k51957
answered Apr 25 '17 at 9:39
Narasimham
20.2k51957
20.2k51957
add a comment |Â
add a comment |Â
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1
The setup is correct. To finish, expand $(e^x+2)^2$ and integrate.
– André Nicolas
Feb 3 '14 at 19:37
1
Note you can pull the $pi$ out of the integral.
– AlexR
Feb 3 '14 at 19:37