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 20 hours ago
David G. Stork
8,96621232
asked 21 hours ago
Renuka
123
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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 20 hours ago
David G. Stork
8,96621232
asked 21 hours ago
Renuka
123
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 20 hours ago
David G. Stork
8,96621232
asked 21 hours ago
Renuka
123
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 20 hours ago
David G. Stork
8,96621232
asked 21 hours ago
Renuka
123
edited 20 hours ago
David G. Stork
8,96621232
asked 21 hours ago
Renuka
123
edited 20 hours ago
David G. Stork
8,96621232
edited 20 hours ago
David G. Stork
8,96621232
edited 20 hours ago
David G. Stork
8,96621232
8,96621232
asked 21 hours ago
Renuka
123
asked 21 hours ago
Renuka
123
asked 21 hours ago
Renuka
123
123
add a comment |
add a comment |
2 Answers
2
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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 21 hours ago
Kavi Rama Murthy
40k31750
add a comment |
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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 21 hours ago
David G. Stork
8,96621232
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 21 hours ago
Kavi Rama Murthy
40k31750
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 21 hours ago
Kavi Rama Murthy
40k31750
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 21 hours ago
Kavi Rama Murthy
40k31750
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 21 hours ago
Kavi Rama Murthy
40k31750
edited 21 hours ago
answered 21 hours ago
Kavi Rama Murthy
40k31750
answered 21 hours ago
Kavi Rama Murthy
40k31750
answered 21 hours ago
Kavi Rama Murthy
40k31750
40k31750
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 21 hours ago
David G. Stork
8,96621232
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 21 hours ago
David G. Stork
8,96621232
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 21 hours ago
David G. Stork
8,96621232
$$intlimits_{x=0}^a intlimits_{y=0}^sqrt{a^2-x^2} (a^2 -x^2) dy dx = frac{3 pi a^4}{16}$$
answered 21 hours ago
David G. Stork
8,96621232
edited 20 hours ago
answered 21 hours ago
David G. Stork
8,96621232
answered 21 hours ago
David G. Stork
8,96621232
answered 21 hours ago
David G. Stork
8,96621232
8,96621232
add a comment |
add a comment |
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