Volume of a solid defined by two curves rotating around $x$-axis.
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
I'm trying to determine the volume of the solid that is obtained by rotating around the $x$-axis the region limited by $y=2x$ and $y=x^2$. I'm trying to follow James Stewart's "Calculus. Single variable" book, but I'm not fully understanding. Is the area given by
$$
A=int_0^2[pi(2x-x^2)^2]dx
$$
Or should I calculate the area between these regions:
$$
A=int(2x-x^2)dx=x^2-fracx^33
$$
And then the volume as:
$$
V=int_0^2[pi(x^2-fracx^33)^2]dx
$$
I'm a bit confused here, any help will be much appreciated.
calculus integration volume
edited Sep 7 at 3:51
zipirovich
10.2k11630
asked Sep 7 at 2:28
Fernando Gómez
1087
add a comment |Â
up vote
0
down vote
favorite
I'm trying to determine the volume of the solid that is obtained by rotating around the $x$-axis the region limited by $y=2x$ and $y=x^2$. I'm trying to follow James Stewart's "Calculus. Single variable" book, but I'm not fully understanding. Is the area given by
$$
A=int_0^2[pi(2x-x^2)^2]dx
$$
Or should I calculate the area between these regions:
$$
A=int(2x-x^2)dx=x^2-fracx^33
$$
And then the volume as:
$$
V=int_0^2[pi(x^2-fracx^33)^2]dx
$$
I'm a bit confused here, any help will be much appreciated.
calculus integration volume
edited Sep 7 at 3:51
zipirovich
10.2k11630
asked Sep 7 at 2:28
Fernando Gómez
1087
1
I don't quite understand why you speak of "area" if the question is to find a volume. That may be part of your confusion. Another hint: there are actually two different, but both perfectly correct, ways to find this volume: using the "cylindrical shells" method or using the "disks and washers" methods. This could be another source of confusion, because the integrals you would set up would be different, and yet both (if done correctly!) would yield the same answer.
– zipirovich
Sep 7 at 3:50
Hi @zipirovich I talk about area because the book I'm following talks about area too. It says that I should calculate the area and then integrate to get the volume. But I'm not sure what area should I take. Quoting from the book (and translating from Spanish as best as I can): "Let S be a solid body that is between $x=a$ and $x=b$. If the area of the transverse section of S over the plane P through x and perpendicular to axis x, is A(x), where A is a continuous function, then the volume is $int_a^bA(x)dx$. " Would that definition fall with any of the two methods you described?
– Fernando Gómez
Sep 7 at 4:10
Just for anybody else having doubts about this, I find the video a day after Norsati gave me the answer, and helped me understand a bit more about these type of volumes. khanacademy.org/math/ap-calculus-ab/…
– Fernando Gómez
Sep 8 at 18:04
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm trying to determine the volume of the solid that is obtained by rotating around the $x$-axis the region limited by $y=2x$ and $y=x^2$. I'm trying to follow James Stewart's "Calculus. Single variable" book, but I'm not fully understanding. Is the area given by
$$
A=int_0^2[pi(2x-x^2)^2]dx
$$
Or should I calculate the area between these regions:
$$
A=int(2x-x^2)dx=x^2-fracx^33
$$
And then the volume as:
$$
V=int_0^2[pi(x^2-fracx^33)^2]dx
$$
I'm a bit confused here, any help will be much appreciated.
calculus integration volume
edited Sep 7 at 3:51
zipirovich
10.2k11630
asked Sep 7 at 2:28
Fernando Gómez
1087
I'm trying to determine the volume of the solid that is obtained by rotating around the $x$-axis the region limited by $y=2x$ and $y=x^2$. I'm trying to follow James Stewart's "Calculus. Single variable" book, but I'm not fully understanding. Is the area given by
$$
A=int_0^2[pi(2x-x^2)^2]dx
$$
Or should I calculate the area between these regions:
$$
A=int(2x-x^2)dx=x^2-fracx^33
$$
And then the volume as:
$$
V=int_0^2[pi(x^2-fracx^33)^2]dx
$$
I'm a bit confused here, any help will be much appreciated.
calculus integration volume
calculus integration volume
edited Sep 7 at 3:51
zipirovich
10.2k11630
asked Sep 7 at 2:28
Fernando Gómez
1087
edited Sep 7 at 3:51
zipirovich
10.2k11630
asked Sep 7 at 2:28
Fernando Gómez
1087
edited Sep 7 at 3:51
zipirovich
10.2k11630
edited Sep 7 at 3:51
zipirovich
10.2k11630
edited Sep 7 at 3:51
zipirovich
10.2k11630
10.2k11630
asked Sep 7 at 2:28
Fernando Gómez
1087
asked Sep 7 at 2:28
Fernando Gómez
1087
asked Sep 7 at 2:28
Fernando Gómez
1087
1087
1
I don't quite understand why you speak of "area" if the question is to find a volume. That may be part of your confusion. Another hint: there are actually two different, but both perfectly correct, ways to find this volume: using the "cylindrical shells" method or using the "disks and washers" methods. This could be another source of confusion, because the integrals you would set up would be different, and yet both (if done correctly!) would yield the same answer.
– zipirovich
Sep 7 at 3:50
Hi @zipirovich I talk about area because the book I'm following talks about area too. It says that I should calculate the area and then integrate to get the volume. But I'm not sure what area should I take. Quoting from the book (and translating from Spanish as best as I can): "Let S be a solid body that is between $x=a$ and $x=b$. If the area of the transverse section of S over the plane P through x and perpendicular to axis x, is A(x), where A is a continuous function, then the volume is $int_a^bA(x)dx$. " Would that definition fall with any of the two methods you described?
– Fernando Gómez
Sep 7 at 4:10
Just for anybody else having doubts about this, I find the video a day after Norsati gave me the answer, and helped me understand a bit more about these type of volumes. khanacademy.org/math/ap-calculus-ab/…
– Fernando Gómez
Sep 8 at 18:04
add a comment |Â
1
I don't quite understand why you speak of "area" if the question is to find a volume. That may be part of your confusion. Another hint: there are actually two different, but both perfectly correct, ways to find this volume: using the "cylindrical shells" method or using the "disks and washers" methods. This could be another source of confusion, because the integrals you would set up would be different, and yet both (if done correctly!) would yield the same answer.
– zipirovich
Sep 7 at 3:50
Hi @zipirovich I talk about area because the book I'm following talks about area too. It says that I should calculate the area and then integrate to get the volume. But I'm not sure what area should I take. Quoting from the book (and translating from Spanish as best as I can): "Let S be a solid body that is between $x=a$ and $x=b$. If the area of the transverse section of S over the plane P through x and perpendicular to axis x, is A(x), where A is a continuous function, then the volume is $int_a^bA(x)dx$. " Would that definition fall with any of the two methods you described?
– Fernando Gómez
Sep 7 at 4:10
Just for anybody else having doubts about this, I find the video a day after Norsati gave me the answer, and helped me understand a bit more about these type of volumes. khanacademy.org/math/ap-calculus-ab/…
– Fernando Gómez
Sep 8 at 18:04
1
1
I don't quite understand why you speak of "area" if the question is to find a volume. That may be part of your confusion. Another hint: there are actually two different, but both perfectly correct, ways to find this volume: using the "cylindrical shells" method or using the "disks and washers" methods. This could be another source of confusion, because the integrals you would set up would be different, and yet both (if done correctly!) would yield the same answer.
– zipirovich
Sep 7 at 3:50
I don't quite understand why you speak of "area" if the question is to find a volume. That may be part of your confusion. Another hint: there are actually two different, but both perfectly correct, ways to find this volume: using the "cylindrical shells" method or using the "disks and washers" methods. This could be another source of confusion, because the integrals you would set up would be different, and yet both (if done correctly!) would yield the same answer.
– zipirovich
Sep 7 at 3:50
Hi @zipirovich I talk about area because the book I'm following talks about area too. It says that I should calculate the area and then integrate to get the volume. But I'm not sure what area should I take. Quoting from the book (and translating from Spanish as best as I can): "Let S be a solid body that is between $x=a$ and $x=b$. If the area of the transverse section of S over the plane P through x and perpendicular to axis x, is A(x), where A is a continuous function, then the volume is $int_a^bA(x)dx$. " Would that definition fall with any of the two methods you described?
– Fernando Gómez
Sep 7 at 4:10
Hi @zipirovich I talk about area because the book I'm following talks about area too. It says that I should calculate the area and then integrate to get the volume. But I'm not sure what area should I take. Quoting from the book (and translating from Spanish as best as I can): "Let S be a solid body that is between $x=a$ and $x=b$. If the area of the transverse section of S over the plane P through x and perpendicular to axis x, is A(x), where A is a continuous function, then the volume is $int_a^bA(x)dx$. " Would that definition fall with any of the two methods you described?
– Fernando Gómez
Sep 7 at 4:10
Just for anybody else having doubts about this, I find the video a day after Norsati gave me the answer, and helped me understand a bit more about these type of volumes. khanacademy.org/math/ap-calculus-ab/…
– Fernando Gómez
Sep 8 at 18:04
Just for anybody else having doubts about this, I find the video a day after Norsati gave me the answer, and helped me understand a bit more about these type of volumes. khanacademy.org/math/ap-calculus-ab/…
– Fernando Gómez
Sep 8 at 18:04
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
For determining the volume of the solid that is obtained by rotating around the $x$-axis limited by $f(x)$ and $g(x)$ where $f(x)geq g(x)$ in $[a,b]$ you have
$$V=piint_a^b f^2(x)-g^2(x) dx$$
here
$$V=piint_0^2 4x^2-x^4 dx$$
answered Sep 7 at 4:27
Nosrati
22.6k61748
Thank you! A quick follow-up: if it were rotated around the y-axis, would the formula change?
– Fernando Gómez
Sep 8 at 17:14
You are welcome. $V=piint_c^d f^2(y)-g^2(y) dy$.
– Nosrati
Sep 8 at 17:17
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
For determining the volume of the solid that is obtained by rotating around the $x$-axis limited by $f(x)$ and $g(x)$ where $f(x)geq g(x)$ in $[a,b]$ you have
$$V=piint_a^b f^2(x)-g^2(x) dx$$
here
$$V=piint_0^2 4x^2-x^4 dx$$
answered Sep 7 at 4:27
Nosrati
22.6k61748
Thank you! A quick follow-up: if it were rotated around the y-axis, would the formula change?
– Fernando Gómez
Sep 8 at 17:14
You are welcome. $V=piint_c^d f^2(y)-g^2(y) dy$.
– Nosrati
Sep 8 at 17:17
add a comment |Â
up vote
2
down vote
accepted
For determining the volume of the solid that is obtained by rotating around the $x$-axis limited by $f(x)$ and $g(x)$ where $f(x)geq g(x)$ in $[a,b]$ you have
$$V=piint_a^b f^2(x)-g^2(x) dx$$
here
$$V=piint_0^2 4x^2-x^4 dx$$
answered Sep 7 at 4:27
Nosrati
22.6k61748
Thank you! A quick follow-up: if it were rotated around the y-axis, would the formula change?
– Fernando Gómez
Sep 8 at 17:14
You are welcome. $V=piint_c^d f^2(y)-g^2(y) dy$.
– Nosrati
Sep 8 at 17:17
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
For determining the volume of the solid that is obtained by rotating around the $x$-axis limited by $f(x)$ and $g(x)$ where $f(x)geq g(x)$ in $[a,b]$ you have
$$V=piint_a^b f^2(x)-g^2(x) dx$$
here
$$V=piint_0^2 4x^2-x^4 dx$$
answered Sep 7 at 4:27
Nosrati
22.6k61748
For determining the volume of the solid that is obtained by rotating around the $x$-axis limited by $f(x)$ and $g(x)$ where $f(x)geq g(x)$ in $[a,b]$ you have
$$V=piint_a^b f^2(x)-g^2(x) dx$$
here
$$V=piint_0^2 4x^2-x^4 dx$$
answered Sep 7 at 4:27
Nosrati
22.6k61748
answered Sep 7 at 4:27
Nosrati
22.6k61748
answered Sep 7 at 4:27
Nosrati
22.6k61748
answered Sep 7 at 4:27
Nosrati
22.6k61748
22.6k61748
Thank you! A quick follow-up: if it were rotated around the y-axis, would the formula change?
– Fernando Gómez
Sep 8 at 17:14
You are welcome. $V=piint_c^d f^2(y)-g^2(y) dy$.
– Nosrati
Sep 8 at 17:17
add a comment |Â
Thank you! A quick follow-up: if it were rotated around the y-axis, would the formula change?
– Fernando Gómez
Sep 8 at 17:14
You are welcome. $V=piint_c^d f^2(y)-g^2(y) dy$.
– Nosrati
Sep 8 at 17:17
Thank you! A quick follow-up: if it were rotated around the y-axis, would the formula change?
– Fernando Gómez
Sep 8 at 17:14
Thank you! A quick follow-up: if it were rotated around the y-axis, would the formula change?
– Fernando Gómez
Sep 8 at 17:14
You are welcome. $V=piint_c^d f^2(y)-g^2(y) dy$.
– Nosrati
Sep 8 at 17:17
You are welcome. $V=piint_c^d f^2(y)-g^2(y) dy$.
– Nosrati
Sep 8 at 17:17
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2908211%2fvolume-of-a-solid-defined-by-two-curves-rotating-around-x-axis%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
1
I don't quite understand why you speak of "area" if the question is to find a volume. That may be part of your confusion. Another hint: there are actually two different, but both perfectly correct, ways to find this volume: using the "cylindrical shells" method or using the "disks and washers" methods. This could be another source of confusion, because the integrals you would set up would be different, and yet both (if done correctly!) would yield the same answer.
– zipirovich
Sep 7 at 3:50
Hi @zipirovich I talk about area because the book I'm following talks about area too. It says that I should calculate the area and then integrate to get the volume. But I'm not sure what area should I take. Quoting from the book (and translating from Spanish as best as I can): "Let S be a solid body that is between $x=a$ and $x=b$. If the area of the transverse section of S over the plane P through x and perpendicular to axis x, is A(x), where A is a continuous function, then the volume is $int_a^bA(x)dx$. " Would that definition fall with any of the two methods you described?
– Fernando Gómez
Sep 7 at 4:10
Just for anybody else having doubts about this, I find the video a day after Norsati gave me the answer, and helped me understand a bit more about these type of volumes. khanacademy.org/math/ap-calculus-ab/…
– Fernando Gómez
Sep 8 at 18:04