find volume using double integral
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Find by double integration volume of sphere $x^2 + y^2 + z^2 = a^2$ cut off by the plane $z = 0$ and the cylinder $x^2 + y^2 = ax$.
I proceded like this :
$x = rcos(theta)$
$x = rsin(theta)$
$r^2 = x^2 + y^2 $
$V = int^ {pi /2}_0int^{acos(theta)}_a {(a^2-r^2)}^{1/2}r~drdtheta$
$V = int^ {pi /2}_0[ {(a^2-r^2)}^{3/2}]^{acos(theta)}_a~dtheta$
following the same path and solving the integral further I got certain ans. but it is wrong. I cant find out what is wrong. Please help.
calculus integration volume
asked Nov 15 at 18:01
atul thakre
316
add a comment |
up vote
0
down vote
favorite
Find by double integration volume of sphere $x^2 + y^2 + z^2 = a^2$ cut off by the plane $z = 0$ and the cylinder $x^2 + y^2 = ax$.
I proceded like this :
$x = rcos(theta)$
$x = rsin(theta)$
$r^2 = x^2 + y^2 $
$V = int^ {pi /2}_0int^{acos(theta)}_a {(a^2-r^2)}^{1/2}r~drdtheta$
$V = int^ {pi /2}_0[ {(a^2-r^2)}^{3/2}]^{acos(theta)}_a~dtheta$
following the same path and solving the integral further I got certain ans. but it is wrong. I cant find out what is wrong. Please help.
calculus integration volume
asked Nov 15 at 18:01
atul thakre
316
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Find by double integration volume of sphere $x^2 + y^2 + z^2 = a^2$ cut off by the plane $z = 0$ and the cylinder $x^2 + y^2 = ax$.
I proceded like this :
$x = rcos(theta)$
$x = rsin(theta)$
$r^2 = x^2 + y^2 $
$V = int^ {pi /2}_0int^{acos(theta)}_a {(a^2-r^2)}^{1/2}r~drdtheta$
$V = int^ {pi /2}_0[ {(a^2-r^2)}^{3/2}]^{acos(theta)}_a~dtheta$
following the same path and solving the integral further I got certain ans. but it is wrong. I cant find out what is wrong. Please help.
calculus integration volume
asked Nov 15 at 18:01
atul thakre
316
Find by double integration volume of sphere $x^2 + y^2 + z^2 = a^2$ cut off by the plane $z = 0$ and the cylinder $x^2 + y^2 = ax$.
I proceded like this :
$x = rcos(theta)$
$x = rsin(theta)$
$r^2 = x^2 + y^2 $
$V = int^ {pi /2}_0int^{acos(theta)}_a {(a^2-r^2)}^{1/2}r~drdtheta$
$V = int^ {pi /2}_0[ {(a^2-r^2)}^{3/2}]^{acos(theta)}_a~dtheta$
following the same path and solving the integral further I got certain ans. but it is wrong. I cant find out what is wrong. Please help.
calculus integration volume
calculus integration volume
asked Nov 15 at 18:01
atul thakre
316
asked Nov 15 at 18:01
atul thakre
316
edited Nov 15 at 18:18
asked Nov 15 at 18:01
atul thakre
316
asked Nov 15 at 18:01
atul thakre
316
asked Nov 15 at 18:01
atul thakre
316
316
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
Your definition of $r$ is missing a square root. But that's probably just a typo. The lower limit of your inside integral should be $0$, not $a$.
But we can't really tell you what you did wrong if you don't post the bit of your work that you think is wrong.
answered Nov 15 at 18:14
B. Goddard
17.9k21340
why the lower limit should be 0 and not a? I thought this for a while earlier
– atul thakre
Nov 15 at 18:18
If the lower limit is $a$ then the lower limit is larger than the upper limit. $r$ ranges from the origin out to the tangent circle $r=acos theta.$
– B. Goddard
Nov 15 at 19:09
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Your definition of $r$ is missing a square root. But that's probably just a typo. The lower limit of your inside integral should be $0$, not $a$.
But we can't really tell you what you did wrong if you don't post the bit of your work that you think is wrong.
answered Nov 15 at 18:14
B. Goddard
17.9k21340
why the lower limit should be 0 and not a? I thought this for a while earlier
– atul thakre
Nov 15 at 18:18
If the lower limit is $a$ then the lower limit is larger than the upper limit. $r$ ranges from the origin out to the tangent circle $r=acos theta.$
– B. Goddard
Nov 15 at 19:09
add a comment |
up vote
0
down vote
Your definition of $r$ is missing a square root. But that's probably just a typo. The lower limit of your inside integral should be $0$, not $a$.
But we can't really tell you what you did wrong if you don't post the bit of your work that you think is wrong.
answered Nov 15 at 18:14
B. Goddard
17.9k21340
why the lower limit should be 0 and not a? I thought this for a while earlier
– atul thakre
Nov 15 at 18:18
If the lower limit is $a$ then the lower limit is larger than the upper limit. $r$ ranges from the origin out to the tangent circle $r=acos theta.$
– B. Goddard
Nov 15 at 19:09
add a comment |
up vote
0
down vote
up vote
0
down vote
Your definition of $r$ is missing a square root. But that's probably just a typo. The lower limit of your inside integral should be $0$, not $a$.
But we can't really tell you what you did wrong if you don't post the bit of your work that you think is wrong.
answered Nov 15 at 18:14
B. Goddard
17.9k21340
Your definition of $r$ is missing a square root. But that's probably just a typo. The lower limit of your inside integral should be $0$, not $a$.
But we can't really tell you what you did wrong if you don't post the bit of your work that you think is wrong.
answered Nov 15 at 18:14
B. Goddard
17.9k21340
answered Nov 15 at 18:14
B. Goddard
17.9k21340
answered Nov 15 at 18:14
B. Goddard
17.9k21340
answered Nov 15 at 18:14
B. Goddard
17.9k21340
17.9k21340
why the lower limit should be 0 and not a? I thought this for a while earlier
– atul thakre
Nov 15 at 18:18
If the lower limit is $a$ then the lower limit is larger than the upper limit. $r$ ranges from the origin out to the tangent circle $r=acos theta.$
– B. Goddard
Nov 15 at 19:09
add a comment |
why the lower limit should be 0 and not a? I thought this for a while earlier
– atul thakre
Nov 15 at 18:18
If the lower limit is $a$ then the lower limit is larger than the upper limit. $r$ ranges from the origin out to the tangent circle $r=acos theta.$
– B. Goddard
Nov 15 at 19:09
why the lower limit should be 0 and not a? I thought this for a while earlier
– atul thakre
Nov 15 at 18:18
why the lower limit should be 0 and not a? I thought this for a while earlier
– atul thakre
Nov 15 at 18:18
If the lower limit is $a$ then the lower limit is larger than the upper limit. $r$ ranges from the origin out to the tangent circle $r=acos theta.$
– B. Goddard
Nov 15 at 19:09
If the lower limit is $a$ then the lower limit is larger than the upper limit. $r$ ranges from the origin out to the tangent circle $r=acos theta.$
– B. Goddard
Nov 15 at 19:09
add a comment |
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