Compute $int int(a^2-x^2) dx dy$ taken over half the circle $x^2+y^2=a^2$ in the positive quadrant.
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
9,28721232
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
9,28721232
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
9,28721232
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
9,28721232
asked Nov 19 at 9:57
Renuka
74
edited Nov 19 at 10:42
David G. Stork
9,28721232
asked Nov 19 at 9:57
Renuka
74
edited Nov 19 at 10:42
David G. Stork
9,28721232
edited Nov 19 at 10:42
David G. Stork
9,28721232
edited Nov 19 at 10:42
David G. Stork
9,28721232
9,28721232
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
2
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
46k31854
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
9,28721232
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
46k31854
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
46k31854
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
46k31854
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
46k31854
edited Nov 19 at 10:08
answered Nov 19 at 10:01
Kavi Rama Murthy
46k31854
answered Nov 19 at 10:01
Kavi Rama Murthy
46k31854
answered Nov 19 at 10:01
Kavi Rama Murthy
46k31854
46k31854
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
9,28721232
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
9,28721232
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
9,28721232
$$intlimits_{x=0}^a intlimits_{y=0}^sqrt{a^2-x^2} (a^2 -x^2) dy dx = frac{3 pi a^4}{16}$$
answered Nov 19 at 10:06
David G. Stork
9,28721232
edited Nov 19 at 10:43
answered Nov 19 at 10:06
David G. Stork
9,28721232
answered Nov 19 at 10:06
David G. Stork
9,28721232
answered Nov 19 at 10:06
David G. Stork
9,28721232
9,28721232
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3004727%2fcompute-int-inta2-x2-dx-dy-taken-over-half-the-circle-x2y2-a2-in%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
