Compute $int int(a^2-x^2) dx dy$ taken over half the circle $x^2+y^2=a^2$ in the positive quadrant.
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Since it is given as positive quadrant, I took the limits as follows,
w.r.t. x from 0 to a & w.r.t y from 0 to root of $$(a^2-x^2)$$.
I proceeded with that & finally substituted $x=asin(theta)$. The answer which I got was $(3(a^4)π)/16$. But the actual answer is $(3a^4)/8$. I don't know where I am making a mistake.
integration
edited Nov 19 at 10:42
David G. Stork
9,25721232
asked Nov 19 at 9:57
Renuka
74
add a comment |
up vote
1
down vote
favorite
Since it is given as positive quadrant, I took the limits as follows,
w.r.t. x from 0 to a & w.r.t y from 0 to root of $$(a^2-x^2)$$.
I proceeded with that & finally substituted $x=asin(theta)$. The answer which I got was $(3(a^4)π)/16$. But the actual answer is $(3a^4)/8$. I don't know where I am making a mistake.
integration
edited Nov 19 at 10:42
David G. Stork
9,25721232
asked Nov 19 at 9:57
Renuka
74
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Since it is given as positive quadrant, I took the limits as follows,
w.r.t. x from 0 to a & w.r.t y from 0 to root of $$(a^2-x^2)$$.
I proceeded with that & finally substituted $x=asin(theta)$. The answer which I got was $(3(a^4)π)/16$. But the actual answer is $(3a^4)/8$. I don't know where I am making a mistake.
integration
edited Nov 19 at 10:42
David G. Stork
9,25721232
asked Nov 19 at 9:57
Renuka
74
Since it is given as positive quadrant, I took the limits as follows,
w.r.t. x from 0 to a & w.r.t y from 0 to root of $$(a^2-x^2)$$.
I proceeded with that & finally substituted $x=asin(theta)$. The answer which I got was $(3(a^4)π)/16$. But the actual answer is $(3a^4)/8$. I don't know where I am making a mistake.
integration
integration
edited Nov 19 at 10:42
David G. Stork
9,25721232
asked Nov 19 at 9:57
Renuka
74
edited Nov 19 at 10:42
David G. Stork
9,25721232
asked Nov 19 at 9:57
Renuka
74
edited Nov 19 at 10:42
David G. Stork
9,25721232
edited Nov 19 at 10:42
David G. Stork
9,25721232
edited Nov 19 at 10:42
David G. Stork
9,25721232
9,25721232
asked Nov 19 at 9:57
Renuka
74
asked Nov 19 at 9:57
Renuka
74
asked Nov 19 at 9:57
Renuka
74
74
add a comment |
add a comment |
2 Answers
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1
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The question refers to a half circle and you are integrating over a quarter of a circle. Perhaps the intended region is ${(x,y):x geq 0, x^{2}+y^{2} leq 1$ in which case $y$ will vary from $-sqrt {a^{2}-x^{2}}$ to $sqrt {a^{2}-x^{2}}$.
answered Nov 19 at 10:01
Kavi Rama Murthy
43.5k31751
add a comment |
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0
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$$intlimits_{x=0}^a intlimits_{y=0}^sqrt{a^2-x^2} (a^2 -x^2) dy dx = frac{3 pi a^4}{16}$$
answered Nov 19 at 10:06
David G. Stork
9,25721232
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
The question refers to a half circle and you are integrating over a quarter of a circle. Perhaps the intended region is ${(x,y):x geq 0, x^{2}+y^{2} leq 1$ in which case $y$ will vary from $-sqrt {a^{2}-x^{2}}$ to $sqrt {a^{2}-x^{2}}$.
answered Nov 19 at 10:01
Kavi Rama Murthy
43.5k31751
add a comment |
up vote
1
down vote
The question refers to a half circle and you are integrating over a quarter of a circle. Perhaps the intended region is ${(x,y):x geq 0, x^{2}+y^{2} leq 1$ in which case $y$ will vary from $-sqrt {a^{2}-x^{2}}$ to $sqrt {a^{2}-x^{2}}$.
answered Nov 19 at 10:01
Kavi Rama Murthy
43.5k31751
add a comment |
up vote
1
down vote
up vote
1
down vote
The question refers to a half circle and you are integrating over a quarter of a circle. Perhaps the intended region is ${(x,y):x geq 0, x^{2}+y^{2} leq 1$ in which case $y$ will vary from $-sqrt {a^{2}-x^{2}}$ to $sqrt {a^{2}-x^{2}}$.
answered Nov 19 at 10:01
Kavi Rama Murthy
43.5k31751
The question refers to a half circle and you are integrating over a quarter of a circle. Perhaps the intended region is ${(x,y):x geq 0, x^{2}+y^{2} leq 1$ in which case $y$ will vary from $-sqrt {a^{2}-x^{2}}$ to $sqrt {a^{2}-x^{2}}$.
answered Nov 19 at 10:01
Kavi Rama Murthy
43.5k31751
edited Nov 19 at 10:08
answered Nov 19 at 10:01
Kavi Rama Murthy
43.5k31751
answered Nov 19 at 10:01
Kavi Rama Murthy
43.5k31751
answered Nov 19 at 10:01
Kavi Rama Murthy
43.5k31751
43.5k31751
add a comment |
add a comment |
up vote
0
down vote
$$intlimits_{x=0}^a intlimits_{y=0}^sqrt{a^2-x^2} (a^2 -x^2) dy dx = frac{3 pi a^4}{16}$$
answered Nov 19 at 10:06
David G. Stork
9,25721232
add a comment |
up vote
0
down vote
$$intlimits_{x=0}^a intlimits_{y=0}^sqrt{a^2-x^2} (a^2 -x^2) dy dx = frac{3 pi a^4}{16}$$
answered Nov 19 at 10:06
David G. Stork
9,25721232
add a comment |
up vote
0
down vote
up vote
0
down vote
$$intlimits_{x=0}^a intlimits_{y=0}^sqrt{a^2-x^2} (a^2 -x^2) dy dx = frac{3 pi a^4}{16}$$
answered Nov 19 at 10:06
David G. Stork
9,25721232
$$intlimits_{x=0}^a intlimits_{y=0}^sqrt{a^2-x^2} (a^2 -x^2) dy dx = frac{3 pi a^4}{16}$$
answered Nov 19 at 10:06
David G. Stork
9,25721232
edited Nov 19 at 10:43
answered Nov 19 at 10:06
David G. Stork
9,25721232
answered Nov 19 at 10:06
David G. Stork
9,25721232
answered Nov 19 at 10:06
David G. Stork
9,25721232
9,25721232
add a comment |
add a comment |
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