Mixing
Volume of intersection of cylinders using polar coordinates
0
$begingroup$
Use a double integral to find the volume of the solid bounded by two surfaces $$x^2+y^2=4,$$ and $$x^2+z^2=4.$$
is it using one way to integrate or there is using separation way?
calculus integration
edited Jan 22 at 11:01
Lee David Chung Lin
4,37031241
asked May 18 '15 at 15:58
Ting Kian RongTing Kian Rong
11
$endgroup$
|
show 2 more comments
0
$begingroup$
Use a double integral to find the volume of the solid bounded by two surfaces $$x^2+y^2=4,$$ and $$x^2+z^2=4.$$
is it using one way to integrate or there is using separation way?
calculus integration
edited Jan 22 at 11:01
Lee David Chung Lin
4,37031241
asked May 18 '15 at 15:58
Ting Kian RongTing Kian Rong
11
$endgroup$
1
$begingroup$
you don't need calculus to solve this problem, see this answer for a related question.
$endgroup$
– achille hui
May 18 '15 at 16:12
$begingroup$
if i want to use the integral way to solve it, can you give me some hint ?
$endgroup$
– Ting Kian Rong
May 19 '15 at 13:58
$begingroup$
You can translate above answer to a triple integral and then integrate over $y$ and $z$ first... i.e. $$int_{-2}^2 left[ int_{-sqrt{4-x^2}}^{sqrt{4-x^2}} int_{-sqrt{4-x^2}}^{sqrt{4-x^2}} dy dz right] dx$$
$endgroup$
– achille hui
May 19 '15 at 14:07
$begingroup$
can i use polar coordinates to do it ? will it be easier to solve right?
$endgroup$
– Ting Kian Rong
May 19 '15 at 15:21
$begingroup$
You can use polar coordinates but it will be a bad choice for this particular problem. The integral over $y$ and $z$ above give you 4(4-x^2), integrating a polynomial is much simpler than integrating any trigonometric function.
$endgroup$
– achille hui
May 19 '15 at 19:51
|
show 2 more comments
0
0
0
$begingroup$
Use a double integral to find the volume of the solid bounded by two surfaces $$x^2+y^2=4,$$ and $$x^2+z^2=4.$$
is it using one way to integrate or there is using separation way?
calculus integration
edited Jan 22 at 11:01
Lee David Chung Lin
4,37031241
asked May 18 '15 at 15:58
Ting Kian RongTing Kian Rong
11
$endgroup$
Use a double integral to find the volume of the solid bounded by two surfaces $$x^2+y^2=4,$$ and $$x^2+z^2=4.$$
is it using one way to integrate or there is using separation way?
calculus integration
calculus integration
edited Jan 22 at 11:01
Lee David Chung Lin
4,37031241
asked May 18 '15 at 15:58
Ting Kian RongTing Kian Rong
11
edited Jan 22 at 11:01
Lee David Chung Lin
4,37031241
asked May 18 '15 at 15:58
Ting Kian RongTing Kian Rong
11
edited Jan 22 at 11:01
Lee David Chung Lin
4,37031241
edited Jan 22 at 11:01
Lee David Chung Lin
4,37031241
edited Jan 22 at 11:01
Lee David Chung Lin
4,37031241
4,37031241
asked May 18 '15 at 15:58
Ting Kian RongTing Kian Rong
11
asked May 18 '15 at 15:58
Ting Kian RongTing Kian Rong
11
asked May 18 '15 at 15:58
Ting Kian RongTing Kian Rong
11
11
1
$begingroup$
you don't need calculus to solve this problem, see this answer for a related question.
$endgroup$
– achille hui
May 18 '15 at 16:12
$begingroup$
if i want to use the integral way to solve it, can you give me some hint ?
$endgroup$
– Ting Kian Rong
May 19 '15 at 13:58
$begingroup$
You can translate above answer to a triple integral and then integrate over $y$ and $z$ first... i.e. $$int_{-2}^2 left[ int_{-sqrt{4-x^2}}^{sqrt{4-x^2}} int_{-sqrt{4-x^2}}^{sqrt{4-x^2}} dy dz right] dx$$
$endgroup$
– achille hui
May 19 '15 at 14:07
$begingroup$
can i use polar coordinates to do it ? will it be easier to solve right?
$endgroup$
– Ting Kian Rong
May 19 '15 at 15:21
$begingroup$
You can use polar coordinates but it will be a bad choice for this particular problem. The integral over $y$ and $z$ above give you 4(4-x^2), integrating a polynomial is much simpler than integrating any trigonometric function.
$endgroup$
– achille hui
May 19 '15 at 19:51
|
show 2 more comments
1
$begingroup$
you don't need calculus to solve this problem, see this answer for a related question.
$endgroup$
– achille hui
May 18 '15 at 16:12
$begingroup$
if i want to use the integral way to solve it, can you give me some hint ?
$endgroup$
– Ting Kian Rong
May 19 '15 at 13:58
$begingroup$
You can translate above answer to a triple integral and then integrate over $y$ and $z$ first... i.e. $$int_{-2}^2 left[ int_{-sqrt{4-x^2}}^{sqrt{4-x^2}} int_{-sqrt{4-x^2}}^{sqrt{4-x^2}} dy dz right] dx$$
$endgroup$
– achille hui
May 19 '15 at 14:07
$begingroup$
can i use polar coordinates to do it ? will it be easier to solve right?
$endgroup$
– Ting Kian Rong
May 19 '15 at 15:21
$begingroup$
You can use polar coordinates but it will be a bad choice for this particular problem. The integral over $y$ and $z$ above give you 4(4-x^2), integrating a polynomial is much simpler than integrating any trigonometric function.
$endgroup$
– achille hui
May 19 '15 at 19:51
1
1
$begingroup$
you don't need calculus to solve this problem, see this answer for a related question.
$endgroup$
– achille hui
May 18 '15 at 16:12
$begingroup$
you don't need calculus to solve this problem, see this answer for a related question.
$endgroup$
– achille hui
May 18 '15 at 16:12
$begingroup$
if i want to use the integral way to solve it, can you give me some hint ?
$endgroup$
– Ting Kian Rong
May 19 '15 at 13:58
$begingroup$
if i want to use the integral way to solve it, can you give me some hint ?
$endgroup$
– Ting Kian Rong
May 19 '15 at 13:58
$begingroup$
You can translate above answer to a triple integral and then integrate over $y$ and $z$ first... i.e. $$int_{-2}^2 left[ int_{-sqrt{4-x^2}}^{sqrt{4-x^2}} int_{-sqrt{4-x^2}}^{sqrt{4-x^2}} dy dz right] dx$$
$endgroup$
– achille hui
May 19 '15 at 14:07
$begingroup$
You can translate above answer to a triple integral and then integrate over $y$ and $z$ first... i.e. $$int_{-2}^2 left[ int_{-sqrt{4-x^2}}^{sqrt{4-x^2}} int_{-sqrt{4-x^2}}^{sqrt{4-x^2}} dy dz right] dx$$
$endgroup$
– achille hui
May 19 '15 at 14:07
$begingroup$
can i use polar coordinates to do it ? will it be easier to solve right?
$endgroup$
– Ting Kian Rong
May 19 '15 at 15:21
$begingroup$
can i use polar coordinates to do it ? will it be easier to solve right?
$endgroup$
– Ting Kian Rong
May 19 '15 at 15:21
$begingroup$
You can use polar coordinates but it will be a bad choice for this particular problem. The integral over $y$ and $z$ above give you 4(4-x^2), integrating a polynomial is much simpler than integrating any trigonometric function.
$endgroup$
– achille hui
May 19 '15 at 19:51
$begingroup$
You can use polar coordinates but it will be a bad choice for this particular problem. The integral over $y$ and $z$ above give you 4(4-x^2), integrating a polynomial is much simpler than integrating any trigonometric function.
$endgroup$
– achille hui
May 19 '15 at 19:51
|
show 2 more comments
0
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1
$begingroup$
you don't need calculus to solve this problem, see this answer for a related question.
$endgroup$
– achille hui
May 18 '15 at 16:12
$begingroup$
if i want to use the integral way to solve it, can you give me some hint ?
$endgroup$
– Ting Kian Rong
May 19 '15 at 13:58
$begingroup$
You can translate above answer to a triple integral and then integrate over $y$ and $z$ first... i.e. $$int_{-2}^2 left[ int_{-sqrt{4-x^2}}^{sqrt{4-x^2}} int_{-sqrt{4-x^2}}^{sqrt{4-x^2}} dy dz right] dx$$
$endgroup$
– achille hui
May 19 '15 at 14:07
$begingroup$
can i use polar coordinates to do it ? will it be easier to solve right?
$endgroup$
– Ting Kian Rong
May 19 '15 at 15:21
$begingroup$
You can use polar coordinates but it will be a bad choice for this particular problem. The integral over $y$ and $z$ above give you 4(4-x^2), integrating a polynomial is much simpler than integrating any trigonometric function.
$endgroup$
– achille hui
May 19 '15 at 19:51