Integral $intlimits_{b-a}^{b+a} frac{sqrt{a^2 - (x-b)^2}}{x} text{d} x$
$begingroup$
I’m trying to solve this integral:
$$intlimits_{b-a}^{b+a} frac{sqrt{a^2 - (x-b)^2}}{x} text{d} x$$
As you can see it isn’t an easy integral to do in cartesian coordinates. However, given the fact that the numerator describes the upper half of the circumference $(x-b)^2 + y^2 = a^2$ this integral could be simpler in polar coordinates.
Yet I’m confused as to how can I use this idea. In polar coordinates $x = r cos (theta)$ so $text{d}x = cos(theta) text{d} r - rsin(theta) text{d} theta$. Do I have to substitute $x$ and $text{d}x$ in order to solve the integral in polar coordinates? Am I even right in thinking that polar coordinates should be used to solve this integral?
calculus definite-integrals
edited Nov 25 '18 at 6:57
Arjang
5,58562363
asked Nov 25 '18 at 6:26
R.MorR.Mor
728
$endgroup$
add a comment |
$begingroup$
I’m trying to solve this integral:
$$intlimits_{b-a}^{b+a} frac{sqrt{a^2 - (x-b)^2}}{x} text{d} x$$
As you can see it isn’t an easy integral to do in cartesian coordinates. However, given the fact that the numerator describes the upper half of the circumference $(x-b)^2 + y^2 = a^2$ this integral could be simpler in polar coordinates.
Yet I’m confused as to how can I use this idea. In polar coordinates $x = r cos (theta)$ so $text{d}x = cos(theta) text{d} r - rsin(theta) text{d} theta$. Do I have to substitute $x$ and $text{d}x$ in order to solve the integral in polar coordinates? Am I even right in thinking that polar coordinates should be used to solve this integral?
calculus definite-integrals
edited Nov 25 '18 at 6:57
Arjang
5,58562363
asked Nov 25 '18 at 6:26
R.MorR.Mor
728
$endgroup$
3
$begingroup$
Substitute $x=acos theta +b.$
$endgroup$
– John_Wick
Nov 25 '18 at 6:51
1
$begingroup$
Note that this substitution is possible because $a cos theta +b$ has a range of $[b-a,b+a]$.
$endgroup$
– Mason
Nov 25 '18 at 7:10
$begingroup$
I’ll try this hint, thank you.
$endgroup$
– R.Mor
Nov 25 '18 at 7:19
$begingroup$
Assuming $a,b in mathbb{R}$ and $b>a$, Mathematica gives: $pi left(b-sqrt{b^2-a^2}right)$.
$endgroup$
– David G. Stork
Nov 25 '18 at 8:33
$begingroup$
I plugged it in Mathematica as well (with the hypothesis you mentioned). However I was trying to solve it and it proved to be lengthy. Maybe in other coordinate system it was easier to do.
$endgroup$
– R.Mor
Nov 25 '18 at 15:04
add a comment |
$begingroup$
I’m trying to solve this integral:
$$intlimits_{b-a}^{b+a} frac{sqrt{a^2 - (x-b)^2}}{x} text{d} x$$
As you can see it isn’t an easy integral to do in cartesian coordinates. However, given the fact that the numerator describes the upper half of the circumference $(x-b)^2 + y^2 = a^2$ this integral could be simpler in polar coordinates.
Yet I’m confused as to how can I use this idea. In polar coordinates $x = r cos (theta)$ so $text{d}x = cos(theta) text{d} r - rsin(theta) text{d} theta$. Do I have to substitute $x$ and $text{d}x$ in order to solve the integral in polar coordinates? Am I even right in thinking that polar coordinates should be used to solve this integral?
calculus definite-integrals
edited Nov 25 '18 at 6:57
Arjang
5,58562363
asked Nov 25 '18 at 6:26
R.MorR.Mor
728
$endgroup$
I’m trying to solve this integral:
$$intlimits_{b-a}^{b+a} frac{sqrt{a^2 - (x-b)^2}}{x} text{d} x$$
As you can see it isn’t an easy integral to do in cartesian coordinates. However, given the fact that the numerator describes the upper half of the circumference $(x-b)^2 + y^2 = a^2$ this integral could be simpler in polar coordinates.
Yet I’m confused as to how can I use this idea. In polar coordinates $x = r cos (theta)$ so $text{d}x = cos(theta) text{d} r - rsin(theta) text{d} theta$. Do I have to substitute $x$ and $text{d}x$ in order to solve the integral in polar coordinates? Am I even right in thinking that polar coordinates should be used to solve this integral?
calculus definite-integrals
calculus definite-integrals
edited Nov 25 '18 at 6:57
Arjang
5,58562363
asked Nov 25 '18 at 6:26
R.MorR.Mor
728
edited Nov 25 '18 at 6:57
Arjang
5,58562363
asked Nov 25 '18 at 6:26
R.MorR.Mor
728
edited Nov 25 '18 at 6:57
Arjang
5,58562363
edited Nov 25 '18 at 6:57
Arjang
5,58562363
edited Nov 25 '18 at 6:57
Arjang
5,58562363
5,58562363
asked Nov 25 '18 at 6:26
R.MorR.Mor
728
asked Nov 25 '18 at 6:26
R.MorR.Mor
728
asked Nov 25 '18 at 6:26
R.MorR.Mor
728
728
3
$begingroup$
Substitute $x=acos theta +b.$
$endgroup$
– John_Wick
Nov 25 '18 at 6:51
1
$begingroup$
Note that this substitution is possible because $a cos theta +b$ has a range of $[b-a,b+a]$.
$endgroup$
– Mason
Nov 25 '18 at 7:10
$begingroup$
I’ll try this hint, thank you.
$endgroup$
– R.Mor
Nov 25 '18 at 7:19
$begingroup$
Assuming $a,b in mathbb{R}$ and $b>a$, Mathematica gives: $pi left(b-sqrt{b^2-a^2}right)$.
$endgroup$
– David G. Stork
Nov 25 '18 at 8:33
$begingroup$
I plugged it in Mathematica as well (with the hypothesis you mentioned). However I was trying to solve it and it proved to be lengthy. Maybe in other coordinate system it was easier to do.
$endgroup$
– R.Mor
Nov 25 '18 at 15:04
add a comment |
3
$begingroup$
Substitute $x=acos theta +b.$
$endgroup$
– John_Wick
Nov 25 '18 at 6:51
1
$begingroup$
Note that this substitution is possible because $a cos theta +b$ has a range of $[b-a,b+a]$.
$endgroup$
– Mason
Nov 25 '18 at 7:10
$begingroup$
I’ll try this hint, thank you.
$endgroup$
– R.Mor
Nov 25 '18 at 7:19
$begingroup$
Assuming $a,b in mathbb{R}$ and $b>a$, Mathematica gives: $pi left(b-sqrt{b^2-a^2}right)$.
$endgroup$
– David G. Stork
Nov 25 '18 at 8:33
$begingroup$
I plugged it in Mathematica as well (with the hypothesis you mentioned). However I was trying to solve it and it proved to be lengthy. Maybe in other coordinate system it was easier to do.
$endgroup$
– R.Mor
Nov 25 '18 at 15:04
3
3
$begingroup$
Substitute $x=acos theta +b.$
$endgroup$
– John_Wick
Nov 25 '18 at 6:51
$begingroup$
Substitute $x=acos theta +b.$
$endgroup$
– John_Wick
Nov 25 '18 at 6:51
1
1
$begingroup$
Note that this substitution is possible because $a cos theta +b$ has a range of $[b-a,b+a]$.
$endgroup$
– Mason
Nov 25 '18 at 7:10
$begingroup$
Note that this substitution is possible because $a cos theta +b$ has a range of $[b-a,b+a]$.
$endgroup$
– Mason
Nov 25 '18 at 7:10
$begingroup$
I’ll try this hint, thank you.
$endgroup$
– R.Mor
Nov 25 '18 at 7:19
$begingroup$
I’ll try this hint, thank you.
$endgroup$
– R.Mor
Nov 25 '18 at 7:19
$begingroup$
Assuming $a,b in mathbb{R}$ and $b>a$, Mathematica gives: $pi left(b-sqrt{b^2-a^2}right)$.
$endgroup$
– David G. Stork
Nov 25 '18 at 8:33
$begingroup$
Assuming $a,b in mathbb{R}$ and $b>a$, Mathematica gives: $pi left(b-sqrt{b^2-a^2}right)$.
$endgroup$
– David G. Stork
Nov 25 '18 at 8:33
$begingroup$
I plugged it in Mathematica as well (with the hypothesis you mentioned). However I was trying to solve it and it proved to be lengthy. Maybe in other coordinate system it was easier to do.
$endgroup$
– R.Mor
Nov 25 '18 at 15:04
$begingroup$
I plugged it in Mathematica as well (with the hypothesis you mentioned). However I was trying to solve it and it proved to be lengthy. Maybe in other coordinate system it was easier to do.
$endgroup$
– R.Mor
Nov 25 '18 at 15:04
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Suppose $b>a>0$. Let $x=b+asin t$ and $u=arctan(frac t2)$ then
begin{eqnarray*}
&&int_{b-a}^{b+a} frac{sqrt{a^2 - (x-b)^2}}{x} text{d} x\
&=&int_{-pi/2}^{pi/2}frac{a^2cos^2 t}{b+asin t}dt\
&=&2a^2int_{-1}^{1}frac{(u^2-1)^2}{(u^2+1)^2(bu^2+2au+b)}du\
&=&2a^2int_{-1}^{1}left[-frac{2u}{(u^2+1)^2}+frac{b}{u^2+1}+frac{a^2-b^2}{a^2(bu^2+2au+b)}right]du\
&=&2a^2left[frac{bpi}{2a^2}+frac{a^2-b^2}{a^2b}int_{-1}^{1}frac{1}{u^2+frac{2a}{b}u+1}duright]\
&=&2left[frac{bpi}{2}+frac{a^2-b^2}{b}int_{-1}^{1}frac{1}{(u+frac{a}{b})^2+frac{b^2-a^2}{b^2}}duright]\
&=&bpi+frac{2(a^2-b^2)}{b}frac{b}{sqrt{b^2-a^2}}arctanleft(frac{b(u+frac{a}{b})}{sqrt{b^2-a^2}}right)bigg|_{-1}^1\
&=&bpi-2sqrt{b^2-a^2}left(arctanleft(frac{b+a}{sqrt{b^2-a^2}}right)-arctanleft(frac{-b+a}{sqrt{b^2-a^2}}right)right)\
&=&bpi-2sqrt{b^2-a^2}cdotfrac{pi}{2}\
&=&pi(b-sqrt{b^2-a^2}).
end{eqnarray*}
answered Nov 26 '18 at 21:35
xpaulxpaul
22.6k24455
$endgroup$
$begingroup$
I arrived at the second integral (the one with the variable of integration $u$), then I got stuck. I didn't think of using partial fractions. Thank you for your answer
$endgroup$
– R.Mor
Nov 27 '18 at 15:28
$begingroup$
You are welcome.
$endgroup$
– xpaul
Nov 27 '18 at 19:39
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
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%2f3012501%2fintegral-int-limits-b-aba-frac-sqrta2-x-b2x-textd-x%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Suppose $b>a>0$. Let $x=b+asin t$ and $u=arctan(frac t2)$ then
begin{eqnarray*}
&&int_{b-a}^{b+a} frac{sqrt{a^2 - (x-b)^2}}{x} text{d} x\
&=&int_{-pi/2}^{pi/2}frac{a^2cos^2 t}{b+asin t}dt\
&=&2a^2int_{-1}^{1}frac{(u^2-1)^2}{(u^2+1)^2(bu^2+2au+b)}du\
&=&2a^2int_{-1}^{1}left[-frac{2u}{(u^2+1)^2}+frac{b}{u^2+1}+frac{a^2-b^2}{a^2(bu^2+2au+b)}right]du\
&=&2a^2left[frac{bpi}{2a^2}+frac{a^2-b^2}{a^2b}int_{-1}^{1}frac{1}{u^2+frac{2a}{b}u+1}duright]\
&=&2left[frac{bpi}{2}+frac{a^2-b^2}{b}int_{-1}^{1}frac{1}{(u+frac{a}{b})^2+frac{b^2-a^2}{b^2}}duright]\
&=&bpi+frac{2(a^2-b^2)}{b}frac{b}{sqrt{b^2-a^2}}arctanleft(frac{b(u+frac{a}{b})}{sqrt{b^2-a^2}}right)bigg|_{-1}^1\
&=&bpi-2sqrt{b^2-a^2}left(arctanleft(frac{b+a}{sqrt{b^2-a^2}}right)-arctanleft(frac{-b+a}{sqrt{b^2-a^2}}right)right)\
&=&bpi-2sqrt{b^2-a^2}cdotfrac{pi}{2}\
&=&pi(b-sqrt{b^2-a^2}).
end{eqnarray*}
answered Nov 26 '18 at 21:35
xpaulxpaul
22.6k24455
$endgroup$
$begingroup$
I arrived at the second integral (the one with the variable of integration $u$), then I got stuck. I didn't think of using partial fractions. Thank you for your answer
$endgroup$
– R.Mor
Nov 27 '18 at 15:28
$begingroup$
You are welcome.
$endgroup$
– xpaul
Nov 27 '18 at 19:39
add a comment |
$begingroup$
Suppose $b>a>0$. Let $x=b+asin t$ and $u=arctan(frac t2)$ then
begin{eqnarray*}
&&int_{b-a}^{b+a} frac{sqrt{a^2 - (x-b)^2}}{x} text{d} x\
&=&int_{-pi/2}^{pi/2}frac{a^2cos^2 t}{b+asin t}dt\
&=&2a^2int_{-1}^{1}frac{(u^2-1)^2}{(u^2+1)^2(bu^2+2au+b)}du\
&=&2a^2int_{-1}^{1}left[-frac{2u}{(u^2+1)^2}+frac{b}{u^2+1}+frac{a^2-b^2}{a^2(bu^2+2au+b)}right]du\
&=&2a^2left[frac{bpi}{2a^2}+frac{a^2-b^2}{a^2b}int_{-1}^{1}frac{1}{u^2+frac{2a}{b}u+1}duright]\
&=&2left[frac{bpi}{2}+frac{a^2-b^2}{b}int_{-1}^{1}frac{1}{(u+frac{a}{b})^2+frac{b^2-a^2}{b^2}}duright]\
&=&bpi+frac{2(a^2-b^2)}{b}frac{b}{sqrt{b^2-a^2}}arctanleft(frac{b(u+frac{a}{b})}{sqrt{b^2-a^2}}right)bigg|_{-1}^1\
&=&bpi-2sqrt{b^2-a^2}left(arctanleft(frac{b+a}{sqrt{b^2-a^2}}right)-arctanleft(frac{-b+a}{sqrt{b^2-a^2}}right)right)\
&=&bpi-2sqrt{b^2-a^2}cdotfrac{pi}{2}\
&=&pi(b-sqrt{b^2-a^2}).
end{eqnarray*}
answered Nov 26 '18 at 21:35
xpaulxpaul
22.6k24455
$endgroup$
$begingroup$
I arrived at the second integral (the one with the variable of integration $u$), then I got stuck. I didn't think of using partial fractions. Thank you for your answer
$endgroup$
– R.Mor
Nov 27 '18 at 15:28
$begingroup$
You are welcome.
$endgroup$
– xpaul
Nov 27 '18 at 19:39
add a comment |
$begingroup$
Suppose $b>a>0$. Let $x=b+asin t$ and $u=arctan(frac t2)$ then
begin{eqnarray*}
&&int_{b-a}^{b+a} frac{sqrt{a^2 - (x-b)^2}}{x} text{d} x\
&=&int_{-pi/2}^{pi/2}frac{a^2cos^2 t}{b+asin t}dt\
&=&2a^2int_{-1}^{1}frac{(u^2-1)^2}{(u^2+1)^2(bu^2+2au+b)}du\
&=&2a^2int_{-1}^{1}left[-frac{2u}{(u^2+1)^2}+frac{b}{u^2+1}+frac{a^2-b^2}{a^2(bu^2+2au+b)}right]du\
&=&2a^2left[frac{bpi}{2a^2}+frac{a^2-b^2}{a^2b}int_{-1}^{1}frac{1}{u^2+frac{2a}{b}u+1}duright]\
&=&2left[frac{bpi}{2}+frac{a^2-b^2}{b}int_{-1}^{1}frac{1}{(u+frac{a}{b})^2+frac{b^2-a^2}{b^2}}duright]\
&=&bpi+frac{2(a^2-b^2)}{b}frac{b}{sqrt{b^2-a^2}}arctanleft(frac{b(u+frac{a}{b})}{sqrt{b^2-a^2}}right)bigg|_{-1}^1\
&=&bpi-2sqrt{b^2-a^2}left(arctanleft(frac{b+a}{sqrt{b^2-a^2}}right)-arctanleft(frac{-b+a}{sqrt{b^2-a^2}}right)right)\
&=&bpi-2sqrt{b^2-a^2}cdotfrac{pi}{2}\
&=&pi(b-sqrt{b^2-a^2}).
end{eqnarray*}
answered Nov 26 '18 at 21:35
xpaulxpaul
22.6k24455
$endgroup$
Suppose $b>a>0$. Let $x=b+asin t$ and $u=arctan(frac t2)$ then
begin{eqnarray*}
&&int_{b-a}^{b+a} frac{sqrt{a^2 - (x-b)^2}}{x} text{d} x\
&=&int_{-pi/2}^{pi/2}frac{a^2cos^2 t}{b+asin t}dt\
&=&2a^2int_{-1}^{1}frac{(u^2-1)^2}{(u^2+1)^2(bu^2+2au+b)}du\
&=&2a^2int_{-1}^{1}left[-frac{2u}{(u^2+1)^2}+frac{b}{u^2+1}+frac{a^2-b^2}{a^2(bu^2+2au+b)}right]du\
&=&2a^2left[frac{bpi}{2a^2}+frac{a^2-b^2}{a^2b}int_{-1}^{1}frac{1}{u^2+frac{2a}{b}u+1}duright]\
&=&2left[frac{bpi}{2}+frac{a^2-b^2}{b}int_{-1}^{1}frac{1}{(u+frac{a}{b})^2+frac{b^2-a^2}{b^2}}duright]\
&=&bpi+frac{2(a^2-b^2)}{b}frac{b}{sqrt{b^2-a^2}}arctanleft(frac{b(u+frac{a}{b})}{sqrt{b^2-a^2}}right)bigg|_{-1}^1\
&=&bpi-2sqrt{b^2-a^2}left(arctanleft(frac{b+a}{sqrt{b^2-a^2}}right)-arctanleft(frac{-b+a}{sqrt{b^2-a^2}}right)right)\
&=&bpi-2sqrt{b^2-a^2}cdotfrac{pi}{2}\
&=&pi(b-sqrt{b^2-a^2}).
end{eqnarray*}
answered Nov 26 '18 at 21:35
xpaulxpaul
22.6k24455
answered Nov 26 '18 at 21:35
xpaulxpaul
22.6k24455
answered Nov 26 '18 at 21:35
xpaulxpaul
22.6k24455
answered Nov 26 '18 at 21:35
xpaulxpaul
22.6k24455
22.6k24455
$begingroup$
I arrived at the second integral (the one with the variable of integration $u$), then I got stuck. I didn't think of using partial fractions. Thank you for your answer
$endgroup$
– R.Mor
Nov 27 '18 at 15:28
$begingroup$
You are welcome.
$endgroup$
– xpaul
Nov 27 '18 at 19:39
add a comment |
$begingroup$
I arrived at the second integral (the one with the variable of integration $u$), then I got stuck. I didn't think of using partial fractions. Thank you for your answer
$endgroup$
– R.Mor
Nov 27 '18 at 15:28
$begingroup$
You are welcome.
$endgroup$
– xpaul
Nov 27 '18 at 19:39
$begingroup$
I arrived at the second integral (the one with the variable of integration $u$), then I got stuck. I didn't think of using partial fractions. Thank you for your answer
$endgroup$
– R.Mor
Nov 27 '18 at 15:28
$begingroup$
I arrived at the second integral (the one with the variable of integration $u$), then I got stuck. I didn't think of using partial fractions. Thank you for your answer
$endgroup$
– R.Mor
Nov 27 '18 at 15:28
$begingroup$
You are welcome.
$endgroup$
– xpaul
Nov 27 '18 at 19:39
$begingroup$
You are welcome.
$endgroup$
– xpaul
Nov 27 '18 at 19:39
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.
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%2f3012501%2fintegral-int-limits-b-aba-frac-sqrta2-x-b2x-textd-x%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
3
$begingroup$
Substitute $x=acos theta +b.$
$endgroup$
– John_Wick
Nov 25 '18 at 6:51
1
$begingroup$
Note that this substitution is possible because $a cos theta +b$ has a range of $[b-a,b+a]$.
$endgroup$
– Mason
Nov 25 '18 at 7:10
$begingroup$
I’ll try this hint, thank you.
$endgroup$
– R.Mor
Nov 25 '18 at 7:19
$begingroup$
Assuming $a,b in mathbb{R}$ and $b>a$, Mathematica gives: $pi left(b-sqrt{b^2-a^2}right)$.
$endgroup$
– David G. Stork
Nov 25 '18 at 8:33
$begingroup$
I plugged it in Mathematica as well (with the hypothesis you mentioned). However I was trying to solve it and it proved to be lengthy. Maybe in other coordinate system it was easier to do.
$endgroup$
– R.Mor
Nov 25 '18 at 15:04