Using disk method in integration
$begingroup$
I need to find the volume of the region defined by the inequalities
$$begin{align*}y & <frac 12\0< y & <1-x^2end{align*}$$revolved around the $x$ axis. I tried separating them and then integrating from zero to $0.707$ for $y<frac 12$ and then from $0.707$ to one for $0<y<1-x^2$ but without a correct result.
integration volume
edited Jan 20 at 19:51
Frank W.
3,6481321
asked Jan 20 at 19:28
Yazan Al-SaifYazan Al-Saif
43
$endgroup$
|
show 2 more comments
$begingroup$
I need to find the volume of the region defined by the inequalities
$$begin{align*}y & <frac 12\0< y & <1-x^2end{align*}$$revolved around the $x$ axis. I tried separating them and then integrating from zero to $0.707$ for $y<frac 12$ and then from $0.707$ to one for $0<y<1-x^2$ but without a correct result.
integration volume
edited Jan 20 at 19:51
Frank W.
3,6481321
asked Jan 20 at 19:28
Yazan Al-SaifYazan Al-Saif
43
$endgroup$
$begingroup$
Welcome to MSE. Please format your question. I don't understand what you are asking. Then tell us what you have tried. Are you stuck on some issue? Does your calculation give the wrong answer? We cannot help if we don't know what you need help with
$endgroup$
– Andrei
Jan 20 at 19:33
$begingroup$
LaTeX formulas need dollar signs around them—only part of your question is being displayed. (I'd edit them in, but am unsure of what's intended.)
$endgroup$
– timtfj
Jan 20 at 19:35
$begingroup$
You can click edit below the question to amend it. While you're editing, a preview should appear below the edit box.
$endgroup$
– timtfj
Jan 20 at 19:37
$begingroup$
Sorry about that. I couldn't make them into formulas but I got them into bold
$endgroup$
– Yazan Al-Saif
Jan 20 at 19:44
$begingroup$
@YazanAl-Saif I've edited the question and embedded your picture. Is this what you want?
$endgroup$
– Frank W.
Jan 20 at 19:51
|
show 2 more comments
$begingroup$
I need to find the volume of the region defined by the inequalities
$$begin{align*}y & <frac 12\0< y & <1-x^2end{align*}$$revolved around the $x$ axis. I tried separating them and then integrating from zero to $0.707$ for $y<frac 12$ and then from $0.707$ to one for $0<y<1-x^2$ but without a correct result.
integration volume
edited Jan 20 at 19:51
Frank W.
3,6481321
asked Jan 20 at 19:28
Yazan Al-SaifYazan Al-Saif
43
$endgroup$
I need to find the volume of the region defined by the inequalities
$$begin{align*}y & <frac 12\0< y & <1-x^2end{align*}$$revolved around the $x$ axis. I tried separating them and then integrating from zero to $0.707$ for $y<frac 12$ and then from $0.707$ to one for $0<y<1-x^2$ but without a correct result.
integration volume
integration volume
edited Jan 20 at 19:51
Frank W.
3,6481321
asked Jan 20 at 19:28
Yazan Al-SaifYazan Al-Saif
43
edited Jan 20 at 19:51
Frank W.
3,6481321
asked Jan 20 at 19:28
Yazan Al-SaifYazan Al-Saif
43
edited Jan 20 at 19:51
Frank W.
3,6481321
edited Jan 20 at 19:51
Frank W.
3,6481321
edited Jan 20 at 19:51
Frank W.
3,6481321
3,6481321
asked Jan 20 at 19:28
Yazan Al-SaifYazan Al-Saif
43
asked Jan 20 at 19:28
Yazan Al-SaifYazan Al-Saif
43
asked Jan 20 at 19:28
Yazan Al-SaifYazan Al-Saif
43
43
$begingroup$
Welcome to MSE. Please format your question. I don't understand what you are asking. Then tell us what you have tried. Are you stuck on some issue? Does your calculation give the wrong answer? We cannot help if we don't know what you need help with
$endgroup$
– Andrei
Jan 20 at 19:33
$begingroup$
LaTeX formulas need dollar signs around them—only part of your question is being displayed. (I'd edit them in, but am unsure of what's intended.)
$endgroup$
– timtfj
Jan 20 at 19:35
$begingroup$
You can click edit below the question to amend it. While you're editing, a preview should appear below the edit box.
$endgroup$
– timtfj
Jan 20 at 19:37
$begingroup$
Sorry about that. I couldn't make them into formulas but I got them into bold
$endgroup$
– Yazan Al-Saif
Jan 20 at 19:44
$begingroup$
@YazanAl-Saif I've edited the question and embedded your picture. Is this what you want?
$endgroup$
– Frank W.
Jan 20 at 19:51
|
show 2 more comments
$begingroup$
Welcome to MSE. Please format your question. I don't understand what you are asking. Then tell us what you have tried. Are you stuck on some issue? Does your calculation give the wrong answer? We cannot help if we don't know what you need help with
$endgroup$
– Andrei
Jan 20 at 19:33
$begingroup$
LaTeX formulas need dollar signs around them—only part of your question is being displayed. (I'd edit them in, but am unsure of what's intended.)
$endgroup$
– timtfj
Jan 20 at 19:35
$begingroup$
You can click edit below the question to amend it. While you're editing, a preview should appear below the edit box.
$endgroup$
– timtfj
Jan 20 at 19:37
$begingroup$
Sorry about that. I couldn't make them into formulas but I got them into bold
$endgroup$
– Yazan Al-Saif
Jan 20 at 19:44
$begingroup$
@YazanAl-Saif I've edited the question and embedded your picture. Is this what you want?
$endgroup$
– Frank W.
Jan 20 at 19:51
$begingroup$
Welcome to MSE. Please format your question. I don't understand what you are asking. Then tell us what you have tried. Are you stuck on some issue? Does your calculation give the wrong answer? We cannot help if we don't know what you need help with
$endgroup$
– Andrei
Jan 20 at 19:33
$begingroup$
Welcome to MSE. Please format your question. I don't understand what you are asking. Then tell us what you have tried. Are you stuck on some issue? Does your calculation give the wrong answer? We cannot help if we don't know what you need help with
$endgroup$
– Andrei
Jan 20 at 19:33
$begingroup$
LaTeX formulas need dollar signs around them—only part of your question is being displayed. (I'd edit them in, but am unsure of what's intended.)
$endgroup$
– timtfj
Jan 20 at 19:35
$begingroup$
LaTeX formulas need dollar signs around them—only part of your question is being displayed. (I'd edit them in, but am unsure of what's intended.)
$endgroup$
– timtfj
Jan 20 at 19:35
$begingroup$
You can click edit below the question to amend it. While you're editing, a preview should appear below the edit box.
$endgroup$
– timtfj
Jan 20 at 19:37
$begingroup$
You can click edit below the question to amend it. While you're editing, a preview should appear below the edit box.
$endgroup$
– timtfj
Jan 20 at 19:37
$begingroup$
Sorry about that. I couldn't make them into formulas but I got them into bold
$endgroup$
– Yazan Al-Saif
Jan 20 at 19:44
$begingroup$
Sorry about that. I couldn't make them into formulas but I got them into bold
$endgroup$
– Yazan Al-Saif
Jan 20 at 19:44
$begingroup$
@YazanAl-Saif I've edited the question and embedded your picture. Is this what you want?
$endgroup$
– Frank W.
Jan 20 at 19:51
$begingroup$
@YazanAl-Saif I've edited the question and embedded your picture. Is this what you want?
$endgroup$
– Frank W.
Jan 20 at 19:51
|
show 2 more comments
2 Answers
2
active
oldest
votes
$begingroup$
The region is simply the area enclosed by the graphs $y=1-x^2$, $y=tfrac 12$, and the $x$ axis. Since the area is symmetric, consider only the region lying in the first quadrant. We'll double our answer to arrive at the area.
First, let's find the intersection of the two functions$$1-x^2=frac 12qquadimpliesqquad x=frac 1{sqrt2}$$
The region to the left of $x$ is simply a rectangle whose area is simply$$A_{text{rectangle}}=frac 1{sqrt2}timesfrac 12=frac 1{2sqrt2}$$The area of the region to the right of $x$ can be done by drawing a "slice" from the function to the $x$ axis. The height of that is simply$$r=1-x^2$$Therefore, rotating it about the $x$ axis and integrating, we get that the area of the right - hand side is simply$$A_{text{right}}=piintlimits_{tfrac 1{sqrt2}}^1mathrm dx,(1-x^2)^2+piintlimits_0^{frac 1{sqrt2}}frac {mathrm dx}4$$
Hence, the total area is$$Acolor{blue}{=2piintlimits_{frac 1{sqrt2}}^1mathrm dx,(1-x^2)^2+2piintlimits_0^{frac 1{sqrt2}}frac {mathrm dx}4}$$
answered Jan 20 at 20:04
Frank W.Frank W.
3,6481321
$endgroup$
$begingroup$
Thank you a lot for the detailed answer
$endgroup$
– Yazan Al-Saif
Jan 20 at 20:10
add a comment |
$begingroup$
For $0to1/sqrt2$, you get$$piint_0^{1/sqrt2}(1/2)^2dx$$For $1/sqrt2to1$, you get$$piint_{1/sqrt2}^1(1-x^2)^2dx$$The final answer is double the sum of the above integrals due to symmetry.
answered Jan 20 at 19:58
Shubham JohriShubham Johri
5,177717
$endgroup$
$begingroup$
I forgot to multiply by 2. Thank you for the answer
$endgroup$
– Yazan Al-Saif
Jan 20 at 20:10
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
$begingroup$
The region is simply the area enclosed by the graphs $y=1-x^2$, $y=tfrac 12$, and the $x$ axis. Since the area is symmetric, consider only the region lying in the first quadrant. We'll double our answer to arrive at the area.
First, let's find the intersection of the two functions$$1-x^2=frac 12qquadimpliesqquad x=frac 1{sqrt2}$$
The region to the left of $x$ is simply a rectangle whose area is simply$$A_{text{rectangle}}=frac 1{sqrt2}timesfrac 12=frac 1{2sqrt2}$$The area of the region to the right of $x$ can be done by drawing a "slice" from the function to the $x$ axis. The height of that is simply$$r=1-x^2$$Therefore, rotating it about the $x$ axis and integrating, we get that the area of the right - hand side is simply$$A_{text{right}}=piintlimits_{tfrac 1{sqrt2}}^1mathrm dx,(1-x^2)^2+piintlimits_0^{frac 1{sqrt2}}frac {mathrm dx}4$$
Hence, the total area is$$Acolor{blue}{=2piintlimits_{frac 1{sqrt2}}^1mathrm dx,(1-x^2)^2+2piintlimits_0^{frac 1{sqrt2}}frac {mathrm dx}4}$$
answered Jan 20 at 20:04
Frank W.Frank W.
3,6481321
$endgroup$
$begingroup$
Thank you a lot for the detailed answer
$endgroup$
– Yazan Al-Saif
Jan 20 at 20:10
add a comment |
$begingroup$
The region is simply the area enclosed by the graphs $y=1-x^2$, $y=tfrac 12$, and the $x$ axis. Since the area is symmetric, consider only the region lying in the first quadrant. We'll double our answer to arrive at the area.
First, let's find the intersection of the two functions$$1-x^2=frac 12qquadimpliesqquad x=frac 1{sqrt2}$$
The region to the left of $x$ is simply a rectangle whose area is simply$$A_{text{rectangle}}=frac 1{sqrt2}timesfrac 12=frac 1{2sqrt2}$$The area of the region to the right of $x$ can be done by drawing a "slice" from the function to the $x$ axis. The height of that is simply$$r=1-x^2$$Therefore, rotating it about the $x$ axis and integrating, we get that the area of the right - hand side is simply$$A_{text{right}}=piintlimits_{tfrac 1{sqrt2}}^1mathrm dx,(1-x^2)^2+piintlimits_0^{frac 1{sqrt2}}frac {mathrm dx}4$$
Hence, the total area is$$Acolor{blue}{=2piintlimits_{frac 1{sqrt2}}^1mathrm dx,(1-x^2)^2+2piintlimits_0^{frac 1{sqrt2}}frac {mathrm dx}4}$$
answered Jan 20 at 20:04
Frank W.Frank W.
3,6481321
$endgroup$
$begingroup$
Thank you a lot for the detailed answer
$endgroup$
– Yazan Al-Saif
Jan 20 at 20:10
add a comment |
$begingroup$
The region is simply the area enclosed by the graphs $y=1-x^2$, $y=tfrac 12$, and the $x$ axis. Since the area is symmetric, consider only the region lying in the first quadrant. We'll double our answer to arrive at the area.
First, let's find the intersection of the two functions$$1-x^2=frac 12qquadimpliesqquad x=frac 1{sqrt2}$$
The region to the left of $x$ is simply a rectangle whose area is simply$$A_{text{rectangle}}=frac 1{sqrt2}timesfrac 12=frac 1{2sqrt2}$$The area of the region to the right of $x$ can be done by drawing a "slice" from the function to the $x$ axis. The height of that is simply$$r=1-x^2$$Therefore, rotating it about the $x$ axis and integrating, we get that the area of the right - hand side is simply$$A_{text{right}}=piintlimits_{tfrac 1{sqrt2}}^1mathrm dx,(1-x^2)^2+piintlimits_0^{frac 1{sqrt2}}frac {mathrm dx}4$$
Hence, the total area is$$Acolor{blue}{=2piintlimits_{frac 1{sqrt2}}^1mathrm dx,(1-x^2)^2+2piintlimits_0^{frac 1{sqrt2}}frac {mathrm dx}4}$$
answered Jan 20 at 20:04
Frank W.Frank W.
3,6481321
$endgroup$
The region is simply the area enclosed by the graphs $y=1-x^2$, $y=tfrac 12$, and the $x$ axis. Since the area is symmetric, consider only the region lying in the first quadrant. We'll double our answer to arrive at the area.
First, let's find the intersection of the two functions$$1-x^2=frac 12qquadimpliesqquad x=frac 1{sqrt2}$$
The region to the left of $x$ is simply a rectangle whose area is simply$$A_{text{rectangle}}=frac 1{sqrt2}timesfrac 12=frac 1{2sqrt2}$$The area of the region to the right of $x$ can be done by drawing a "slice" from the function to the $x$ axis. The height of that is simply$$r=1-x^2$$Therefore, rotating it about the $x$ axis and integrating, we get that the area of the right - hand side is simply$$A_{text{right}}=piintlimits_{tfrac 1{sqrt2}}^1mathrm dx,(1-x^2)^2+piintlimits_0^{frac 1{sqrt2}}frac {mathrm dx}4$$
Hence, the total area is$$Acolor{blue}{=2piintlimits_{frac 1{sqrt2}}^1mathrm dx,(1-x^2)^2+2piintlimits_0^{frac 1{sqrt2}}frac {mathrm dx}4}$$
answered Jan 20 at 20:04
Frank W.Frank W.
3,6481321
answered Jan 20 at 20:04
Frank W.Frank W.
3,6481321
answered Jan 20 at 20:04
Frank W.Frank W.
3,6481321
answered Jan 20 at 20:04
Frank W.Frank W.
3,6481321
3,6481321
$begingroup$
Thank you a lot for the detailed answer
$endgroup$
– Yazan Al-Saif
Jan 20 at 20:10
add a comment |
$begingroup$
Thank you a lot for the detailed answer
$endgroup$
– Yazan Al-Saif
Jan 20 at 20:10
$begingroup$
Thank you a lot for the detailed answer
$endgroup$
– Yazan Al-Saif
Jan 20 at 20:10
$begingroup$
Thank you a lot for the detailed answer
$endgroup$
– Yazan Al-Saif
Jan 20 at 20:10
add a comment |
$begingroup$
For $0to1/sqrt2$, you get$$piint_0^{1/sqrt2}(1/2)^2dx$$For $1/sqrt2to1$, you get$$piint_{1/sqrt2}^1(1-x^2)^2dx$$The final answer is double the sum of the above integrals due to symmetry.
answered Jan 20 at 19:58
Shubham JohriShubham Johri
5,177717
$endgroup$
$begingroup$
I forgot to multiply by 2. Thank you for the answer
$endgroup$
– Yazan Al-Saif
Jan 20 at 20:10
add a comment |
$begingroup$
For $0to1/sqrt2$, you get$$piint_0^{1/sqrt2}(1/2)^2dx$$For $1/sqrt2to1$, you get$$piint_{1/sqrt2}^1(1-x^2)^2dx$$The final answer is double the sum of the above integrals due to symmetry.
answered Jan 20 at 19:58
Shubham JohriShubham Johri
5,177717
$endgroup$
$begingroup$
I forgot to multiply by 2. Thank you for the answer
$endgroup$
– Yazan Al-Saif
Jan 20 at 20:10
add a comment |
$begingroup$
For $0to1/sqrt2$, you get$$piint_0^{1/sqrt2}(1/2)^2dx$$For $1/sqrt2to1$, you get$$piint_{1/sqrt2}^1(1-x^2)^2dx$$The final answer is double the sum of the above integrals due to symmetry.
answered Jan 20 at 19:58
Shubham JohriShubham Johri
5,177717
$endgroup$
For $0to1/sqrt2$, you get$$piint_0^{1/sqrt2}(1/2)^2dx$$For $1/sqrt2to1$, you get$$piint_{1/sqrt2}^1(1-x^2)^2dx$$The final answer is double the sum of the above integrals due to symmetry.
answered Jan 20 at 19:58
Shubham JohriShubham Johri
5,177717
answered Jan 20 at 19:58
Shubham JohriShubham Johri
5,177717
answered Jan 20 at 19:58
Shubham JohriShubham Johri
5,177717
answered Jan 20 at 19:58
Shubham JohriShubham Johri
5,177717
5,177717
$begingroup$
I forgot to multiply by 2. Thank you for the answer
$endgroup$
– Yazan Al-Saif
Jan 20 at 20:10
add a comment |
$begingroup$
I forgot to multiply by 2. Thank you for the answer
$endgroup$
– Yazan Al-Saif
Jan 20 at 20:10
$begingroup$
I forgot to multiply by 2. Thank you for the answer
$endgroup$
– Yazan Al-Saif
Jan 20 at 20:10
$begingroup$
I forgot to multiply by 2. Thank you for the answer
$endgroup$
– Yazan Al-Saif
Jan 20 at 20:10
add a comment |
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$begingroup$
Welcome to MSE. Please format your question. I don't understand what you are asking. Then tell us what you have tried. Are you stuck on some issue? Does your calculation give the wrong answer? We cannot help if we don't know what you need help with
$endgroup$
– Andrei
Jan 20 at 19:33
$begingroup$
LaTeX formulas need dollar signs around them—only part of your question is being displayed. (I'd edit them in, but am unsure of what's intended.)
$endgroup$
– timtfj
Jan 20 at 19:35
$begingroup$
You can click edit below the question to amend it. While you're editing, a preview should appear below the edit box.
$endgroup$
– timtfj
Jan 20 at 19:37
$begingroup$
Sorry about that. I couldn't make them into formulas but I got them into bold
$endgroup$
– Yazan Al-Saif
Jan 20 at 19:44
$begingroup$
@YazanAl-Saif I've edited the question and embedded your picture. Is this what you want?
$endgroup$
– Frank W.
Jan 20 at 19:51