Mixing
![How much had really been ordered? [closed]](https://lh6.googleusercontent.com/-d_yOqCzrknI/AAAAAAAAAAI/AAAAAAAAIrw/IWUHx2BqztA/s72-c/photo.jpg?sz=32)
Solid of revolution using washer method(Gives negative answer)
$begingroup$
My teacher even knew that the answer should not be negative but it turned out to be negative. The given was y=x^2, y=4x-x^2, revolving about the y-axis. Here are some of the solution presented, I hope someone could help me find where I went wrong while solving. Thanks in advance. My teacher can’t even find the error that I have committed.
Edit: As I did the right - left thingy, the answer was -112/3(pi) cu. units.
V=(pi)[(sqrt(y))^2-(sqrt(4-y)+2)^2)]dy
V=(pi)[2y-4sqrt(4-y)-8]dy
V=(pi)[y^2+8/3(4-y)^3/2-8y] limit from 0 to 4 <——already integrated
Applying the limits
V=(pi)[16+8/3(0-8)]
V=-112/3(pi) cu. units. Is there anything done wrong? I hope someone can spot what is wrong. I really need the correct answer utgently. Thanks for those who can help.
area solid-of-revolution
asked Jan 31 at 0:07
Bido262Bido262
103
$endgroup$
add a comment |
$begingroup$
My teacher even knew that the answer should not be negative but it turned out to be negative. The given was y=x^2, y=4x-x^2, revolving about the y-axis. Here are some of the solution presented, I hope someone could help me find where I went wrong while solving. Thanks in advance. My teacher can’t even find the error that I have committed.
Edit: As I did the right - left thingy, the answer was -112/3(pi) cu. units.
V=(pi)[(sqrt(y))^2-(sqrt(4-y)+2)^2)]dy
V=(pi)[2y-4sqrt(4-y)-8]dy
V=(pi)[y^2+8/3(4-y)^3/2-8y] limit from 0 to 4 <——already integrated
Applying the limits
V=(pi)[16+8/3(0-8)]
V=-112/3(pi) cu. units. Is there anything done wrong? I hope someone can spot what is wrong. I really need the correct answer utgently. Thanks for those who can help.
area solid-of-revolution
asked Jan 31 at 0:07
Bido262Bido262
103
$endgroup$
$begingroup$
Are you sure you've included your work in the post? I'm not seeing it
$endgroup$
– Alex
Jan 31 at 0:18
$begingroup$
Sorry if you’ve not seen it. Never knew that my photo was not uploaded. Now you could probably check again. Sorry for the incovenience. The graph is the same as what the decaf-math answered.
$endgroup$
– Bido262
Jan 31 at 15:52
add a comment |
$begingroup$
My teacher even knew that the answer should not be negative but it turned out to be negative. The given was y=x^2, y=4x-x^2, revolving about the y-axis. Here are some of the solution presented, I hope someone could help me find where I went wrong while solving. Thanks in advance. My teacher can’t even find the error that I have committed.
Edit: As I did the right - left thingy, the answer was -112/3(pi) cu. units.
V=(pi)[(sqrt(y))^2-(sqrt(4-y)+2)^2)]dy
V=(pi)[2y-4sqrt(4-y)-8]dy
V=(pi)[y^2+8/3(4-y)^3/2-8y] limit from 0 to 4 <——already integrated
Applying the limits
V=(pi)[16+8/3(0-8)]
V=-112/3(pi) cu. units. Is there anything done wrong? I hope someone can spot what is wrong. I really need the correct answer utgently. Thanks for those who can help.
area solid-of-revolution
asked Jan 31 at 0:07
Bido262Bido262
103
$endgroup$
My teacher even knew that the answer should not be negative but it turned out to be negative. The given was y=x^2, y=4x-x^2, revolving about the y-axis. Here are some of the solution presented, I hope someone could help me find where I went wrong while solving. Thanks in advance. My teacher can’t even find the error that I have committed.
Edit: As I did the right - left thingy, the answer was -112/3(pi) cu. units.
V=(pi)[(sqrt(y))^2-(sqrt(4-y)+2)^2)]dy
V=(pi)[2y-4sqrt(4-y)-8]dy
V=(pi)[y^2+8/3(4-y)^3/2-8y] limit from 0 to 4 <——already integrated
Applying the limits
V=(pi)[16+8/3(0-8)]
V=-112/3(pi) cu. units. Is there anything done wrong? I hope someone can spot what is wrong. I really need the correct answer utgently. Thanks for those who can help.
area solid-of-revolution
area solid-of-revolution
asked Jan 31 at 0:07
Bido262Bido262
103
asked Jan 31 at 0:07
Bido262Bido262
103
edited Jan 31 at 15:49
Bido262
asked Jan 31 at 0:07
Bido262Bido262
103
asked Jan 31 at 0:07
Bido262Bido262
103
asked Jan 31 at 0:07
Bido262Bido262
103
103
$begingroup$
Are you sure you've included your work in the post? I'm not seeing it
$endgroup$
– Alex
Jan 31 at 0:18
$begingroup$
Sorry if you’ve not seen it. Never knew that my photo was not uploaded. Now you could probably check again. Sorry for the incovenience. The graph is the same as what the decaf-math answered.
$endgroup$
– Bido262
Jan 31 at 15:52
add a comment |
$begingroup$
Are you sure you've included your work in the post? I'm not seeing it
$endgroup$
– Alex
Jan 31 at 0:18
$begingroup$
Sorry if you’ve not seen it. Never knew that my photo was not uploaded. Now you could probably check again. Sorry for the incovenience. The graph is the same as what the decaf-math answered.
$endgroup$
– Bido262
Jan 31 at 15:52
$begingroup$
Are you sure you've included your work in the post? I'm not seeing it
$endgroup$
– Alex
Jan 31 at 0:18
$begingroup$
Are you sure you've included your work in the post? I'm not seeing it
$endgroup$
– Alex
Jan 31 at 0:18
$begingroup$
Sorry if you’ve not seen it. Never knew that my photo was not uploaded. Now you could probably check again. Sorry for the incovenience. The graph is the same as what the decaf-math answered.
$endgroup$
– Bido262
Jan 31 at 15:52
$begingroup$
Sorry if you’ve not seen it. Never knew that my photo was not uploaded. Now you could probably check again. Sorry for the incovenience. The graph is the same as what the decaf-math answered.
$endgroup$
– Bido262
Jan 31 at 15:52
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint: if you plan on using the $y$-axis as your axis of rotation for the washer method, then you are going to need to integrate with respect to $y$, and you need to solve each of the equations in terms of $x$: $$y = x^2 implies pmsqrt y = x implies sqrt y = x checkmark qquadtext{(Since we need the right-half)}$$
However, the other equation is not so straightforward. You will need to complete the square to help eliminate one of the $x$'s: $$begin{align}y &= 4x-x^2\ &= -(x^2-4x + 4 - 4) \ &= -(x-2)^2 + 4 \ implies y- 4 &= -(x-2)^2\ pmsqrt{4-y} &= x-2 \ implies pmsqrt{4-y}+2 &= x\ implies -sqrt{4-y}+2 &= x. qquadtext{(Since we need the left-half)}end{align}$$
Notice how $sqrt y = x$ is the rightmost graph and $-sqrt{4-y}+2 = x$ is the leftmost. You will need to do right - left in your radius function when computing the integral.
answered Jan 31 at 0:40
Decaf-MathDecaf-Math
3,422926
$endgroup$
$begingroup$
Yea. I got the same constraints. Got same solutions but my answer turns out negative. Go check my post again and look if I did something wrong. Ps: I edited the post.
$endgroup$
– Bido262
Jan 31 at 15:53
1
$begingroup$
@Bido262, one thing that I notice immediately is that you use the wrong half of one of the functions. You wrote $$V=piint_0^4[(sqrt y )^2-(sqrt{4-y}+2)^2)]dy,$$ but it should be $$V=piint_0^4left[(sqrt y)^2 - (2-sqrt{4-y})^2right]dy.$$ If you compute that instead, you get a positive answer.
$endgroup$
– Decaf-Math
Jan 31 at 16:02
$begingroup$
Omg. Finally. Thanks mate, will be presenting this damn problem tomorrow. The answer turned out to be 16/3(pi) cu. units. Pls correct me if am wrong.
$endgroup$
– Bido262
Jan 31 at 16:18
$begingroup$
@Bido262 $displaystyle {16piover3}$ is indeed the answer I received.
$endgroup$
– Decaf-Math
Jan 31 at 16:22
$begingroup$
How to mark this question as anwered?
$endgroup$
– Bido262
Jan 31 at 16:23
|
show 1 more comment
Your Answer
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1 Answer
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$begingroup$
Hint: if you plan on using the $y$-axis as your axis of rotation for the washer method, then you are going to need to integrate with respect to $y$, and you need to solve each of the equations in terms of $x$: $$y = x^2 implies pmsqrt y = x implies sqrt y = x checkmark qquadtext{(Since we need the right-half)}$$
However, the other equation is not so straightforward. You will need to complete the square to help eliminate one of the $x$'s: $$begin{align}y &= 4x-x^2\ &= -(x^2-4x + 4 - 4) \ &= -(x-2)^2 + 4 \ implies y- 4 &= -(x-2)^2\ pmsqrt{4-y} &= x-2 \ implies pmsqrt{4-y}+2 &= x\ implies -sqrt{4-y}+2 &= x. qquadtext{(Since we need the left-half)}end{align}$$
Notice how $sqrt y = x$ is the rightmost graph and $-sqrt{4-y}+2 = x$ is the leftmost. You will need to do right - left in your radius function when computing the integral.
answered Jan 31 at 0:40
Decaf-MathDecaf-Math
3,422926
$endgroup$
$begingroup$
Yea. I got the same constraints. Got same solutions but my answer turns out negative. Go check my post again and look if I did something wrong. Ps: I edited the post.
$endgroup$
– Bido262
Jan 31 at 15:53
1
$begingroup$
@Bido262, one thing that I notice immediately is that you use the wrong half of one of the functions. You wrote $$V=piint_0^4[(sqrt y )^2-(sqrt{4-y}+2)^2)]dy,$$ but it should be $$V=piint_0^4left[(sqrt y)^2 - (2-sqrt{4-y})^2right]dy.$$ If you compute that instead, you get a positive answer.
$endgroup$
– Decaf-Math
Jan 31 at 16:02
$begingroup$
Omg. Finally. Thanks mate, will be presenting this damn problem tomorrow. The answer turned out to be 16/3(pi) cu. units. Pls correct me if am wrong.
$endgroup$
– Bido262
Jan 31 at 16:18
$begingroup$
@Bido262 $displaystyle {16piover3}$ is indeed the answer I received.
$endgroup$
– Decaf-Math
Jan 31 at 16:22
$begingroup$
How to mark this question as anwered?
$endgroup$
– Bido262
Jan 31 at 16:23
|
show 1 more comment
$begingroup$
Hint: if you plan on using the $y$-axis as your axis of rotation for the washer method, then you are going to need to integrate with respect to $y$, and you need to solve each of the equations in terms of $x$: $$y = x^2 implies pmsqrt y = x implies sqrt y = x checkmark qquadtext{(Since we need the right-half)}$$
However, the other equation is not so straightforward. You will need to complete the square to help eliminate one of the $x$'s: $$begin{align}y &= 4x-x^2\ &= -(x^2-4x + 4 - 4) \ &= -(x-2)^2 + 4 \ implies y- 4 &= -(x-2)^2\ pmsqrt{4-y} &= x-2 \ implies pmsqrt{4-y}+2 &= x\ implies -sqrt{4-y}+2 &= x. qquadtext{(Since we need the left-half)}end{align}$$
Notice how $sqrt y = x$ is the rightmost graph and $-sqrt{4-y}+2 = x$ is the leftmost. You will need to do right - left in your radius function when computing the integral.
answered Jan 31 at 0:40
Decaf-MathDecaf-Math
3,422926
$endgroup$
$begingroup$
Yea. I got the same constraints. Got same solutions but my answer turns out negative. Go check my post again and look if I did something wrong. Ps: I edited the post.
$endgroup$
– Bido262
Jan 31 at 15:53
1
$begingroup$
@Bido262, one thing that I notice immediately is that you use the wrong half of one of the functions. You wrote $$V=piint_0^4[(sqrt y )^2-(sqrt{4-y}+2)^2)]dy,$$ but it should be $$V=piint_0^4left[(sqrt y)^2 - (2-sqrt{4-y})^2right]dy.$$ If you compute that instead, you get a positive answer.
$endgroup$
– Decaf-Math
Jan 31 at 16:02
$begingroup$
Omg. Finally. Thanks mate, will be presenting this damn problem tomorrow. The answer turned out to be 16/3(pi) cu. units. Pls correct me if am wrong.
$endgroup$
– Bido262
Jan 31 at 16:18
$begingroup$
@Bido262 $displaystyle {16piover3}$ is indeed the answer I received.
$endgroup$
– Decaf-Math
Jan 31 at 16:22
$begingroup$
How to mark this question as anwered?
$endgroup$
– Bido262
Jan 31 at 16:23
|
show 1 more comment
$begingroup$
Hint: if you plan on using the $y$-axis as your axis of rotation for the washer method, then you are going to need to integrate with respect to $y$, and you need to solve each of the equations in terms of $x$: $$y = x^2 implies pmsqrt y = x implies sqrt y = x checkmark qquadtext{(Since we need the right-half)}$$
However, the other equation is not so straightforward. You will need to complete the square to help eliminate one of the $x$'s: $$begin{align}y &= 4x-x^2\ &= -(x^2-4x + 4 - 4) \ &= -(x-2)^2 + 4 \ implies y- 4 &= -(x-2)^2\ pmsqrt{4-y} &= x-2 \ implies pmsqrt{4-y}+2 &= x\ implies -sqrt{4-y}+2 &= x. qquadtext{(Since we need the left-half)}end{align}$$
Notice how $sqrt y = x$ is the rightmost graph and $-sqrt{4-y}+2 = x$ is the leftmost. You will need to do right - left in your radius function when computing the integral.
answered Jan 31 at 0:40
Decaf-MathDecaf-Math
3,422926
$endgroup$
Hint: if you plan on using the $y$-axis as your axis of rotation for the washer method, then you are going to need to integrate with respect to $y$, and you need to solve each of the equations in terms of $x$: $$y = x^2 implies pmsqrt y = x implies sqrt y = x checkmark qquadtext{(Since we need the right-half)}$$
However, the other equation is not so straightforward. You will need to complete the square to help eliminate one of the $x$'s: $$begin{align}y &= 4x-x^2\ &= -(x^2-4x + 4 - 4) \ &= -(x-2)^2 + 4 \ implies y- 4 &= -(x-2)^2\ pmsqrt{4-y} &= x-2 \ implies pmsqrt{4-y}+2 &= x\ implies -sqrt{4-y}+2 &= x. qquadtext{(Since we need the left-half)}end{align}$$
Notice how $sqrt y = x$ is the rightmost graph and $-sqrt{4-y}+2 = x$ is the leftmost. You will need to do right - left in your radius function when computing the integral.
answered Jan 31 at 0:40
Decaf-MathDecaf-Math
3,422926
answered Jan 31 at 0:40
Decaf-MathDecaf-Math
3,422926
answered Jan 31 at 0:40
Decaf-MathDecaf-Math
3,422926
answered Jan 31 at 0:40
Decaf-MathDecaf-Math
3,422926
3,422926
$begingroup$
Yea. I got the same constraints. Got same solutions but my answer turns out negative. Go check my post again and look if I did something wrong. Ps: I edited the post.
$endgroup$
– Bido262
Jan 31 at 15:53
1
$begingroup$
@Bido262, one thing that I notice immediately is that you use the wrong half of one of the functions. You wrote $$V=piint_0^4[(sqrt y )^2-(sqrt{4-y}+2)^2)]dy,$$ but it should be $$V=piint_0^4left[(sqrt y)^2 - (2-sqrt{4-y})^2right]dy.$$ If you compute that instead, you get a positive answer.
$endgroup$
– Decaf-Math
Jan 31 at 16:02
$begingroup$
Omg. Finally. Thanks mate, will be presenting this damn problem tomorrow. The answer turned out to be 16/3(pi) cu. units. Pls correct me if am wrong.
$endgroup$
– Bido262
Jan 31 at 16:18
$begingroup$
@Bido262 $displaystyle {16piover3}$ is indeed the answer I received.
$endgroup$
– Decaf-Math
Jan 31 at 16:22
$begingroup$
How to mark this question as anwered?
$endgroup$
– Bido262
Jan 31 at 16:23
|
show 1 more comment
$begingroup$
Yea. I got the same constraints. Got same solutions but my answer turns out negative. Go check my post again and look if I did something wrong. Ps: I edited the post.
$endgroup$
– Bido262
Jan 31 at 15:53
1
$begingroup$
@Bido262, one thing that I notice immediately is that you use the wrong half of one of the functions. You wrote $$V=piint_0^4[(sqrt y )^2-(sqrt{4-y}+2)^2)]dy,$$ but it should be $$V=piint_0^4left[(sqrt y)^2 - (2-sqrt{4-y})^2right]dy.$$ If you compute that instead, you get a positive answer.
$endgroup$
– Decaf-Math
Jan 31 at 16:02
$begingroup$
Omg. Finally. Thanks mate, will be presenting this damn problem tomorrow. The answer turned out to be 16/3(pi) cu. units. Pls correct me if am wrong.
$endgroup$
– Bido262
Jan 31 at 16:18
$begingroup$
@Bido262 $displaystyle {16piover3}$ is indeed the answer I received.
$endgroup$
– Decaf-Math
Jan 31 at 16:22
$begingroup$
How to mark this question as anwered?
$endgroup$
– Bido262
Jan 31 at 16:23
$begingroup$
Yea. I got the same constraints. Got same solutions but my answer turns out negative. Go check my post again and look if I did something wrong. Ps: I edited the post.
$endgroup$
– Bido262
Jan 31 at 15:53
$begingroup$
Yea. I got the same constraints. Got same solutions but my answer turns out negative. Go check my post again and look if I did something wrong. Ps: I edited the post.
$endgroup$
– Bido262
Jan 31 at 15:53
1
1
$begingroup$
@Bido262, one thing that I notice immediately is that you use the wrong half of one of the functions. You wrote $$V=piint_0^4[(sqrt y )^2-(sqrt{4-y}+2)^2)]dy,$$ but it should be $$V=piint_0^4left[(sqrt y)^2 - (2-sqrt{4-y})^2right]dy.$$ If you compute that instead, you get a positive answer.
$endgroup$
– Decaf-Math
Jan 31 at 16:02
$begingroup$
@Bido262, one thing that I notice immediately is that you use the wrong half of one of the functions. You wrote $$V=piint_0^4[(sqrt y )^2-(sqrt{4-y}+2)^2)]dy,$$ but it should be $$V=piint_0^4left[(sqrt y)^2 - (2-sqrt{4-y})^2right]dy.$$ If you compute that instead, you get a positive answer.
$endgroup$
– Decaf-Math
Jan 31 at 16:02
$begingroup$
Omg. Finally. Thanks mate, will be presenting this damn problem tomorrow. The answer turned out to be 16/3(pi) cu. units. Pls correct me if am wrong.
$endgroup$
– Bido262
Jan 31 at 16:18
$begingroup$
Omg. Finally. Thanks mate, will be presenting this damn problem tomorrow. The answer turned out to be 16/3(pi) cu. units. Pls correct me if am wrong.
$endgroup$
– Bido262
Jan 31 at 16:18
$begingroup$
@Bido262 $displaystyle {16piover3}$ is indeed the answer I received.
$endgroup$
– Decaf-Math
Jan 31 at 16:22
$begingroup$
@Bido262 $displaystyle {16piover3}$ is indeed the answer I received.
$endgroup$
– Decaf-Math
Jan 31 at 16:22
$begingroup$
How to mark this question as anwered?
$endgroup$
– Bido262
Jan 31 at 16:23
$begingroup$
How to mark this question as anwered?
$endgroup$
– Bido262
Jan 31 at 16:23
|
show 1 more comment
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$begingroup$
Are you sure you've included your work in the post? I'm not seeing it
$endgroup$
– Alex
Jan 31 at 0:18
$begingroup$
Sorry if you’ve not seen it. Never knew that my photo was not uploaded. Now you could probably check again. Sorry for the incovenience. The graph is the same as what the decaf-math answered.
$endgroup$
– Bido262
Jan 31 at 15:52