How do I change $\int_{0}^{1}{\int_{0}^{\sqrt{1-{{x}^{2}}}}{\int_{\sqrt{{{x}^{2}}+{{y}^{2}}}}^{\sqrt{2-{{x}^{2}}-{{y}^{2}}}}{xydzdydx}}}$ to cylindrical or spherical coordinates?
Answer
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Hint: In this question, we have to convert the given definite integral into cylindrical coordinates or the spherical coordinates. Thus, we start our problem by solving the limit z of the given integral. After that, we will find the value of $\rho $ from the second limit of the z. After that, we will find the value of z by substituting the value of the first limit of z in the second limit of z. Then, we will make the necessary calculations and then we will find the value of $\phi $ and $\theta $ from the conversion of z and from the equation of sphere. In the end, we will put the new limits in the integral and the x, y conversions, to get the required result for the problem.
Complete step-by-step solution:
According to the problem, we have to find the cylindrical or spherical coordinates from a definite integral.
Thus, we will apply the polar form to get the solution.
The integral given to us is $\int_{0}^{1}{\int_{0}^{\sqrt{1-{{x}^{2}}}}{\int_{\sqrt{{{x}^{2}}+{{y}^{2}}}}^{\sqrt{2-{{x}^{2}}-{{y}^{2}}}}{xydzdydx}}}$ --------- (1)
So, from equation (1), we get the value of the limit equals to
$\sqrt{{{x}^{2}}+{{y}^{2}}}\le z\le \sqrt{2-{{x}^{2}}-{{y}^{2}}}$
$0\le y\le \sqrt{1-{{x}^{2}}}$
$0\le x\le 1$
Now, we will first solve the z-domain, that is
$z=\sqrt{{{x}^{2}}+{{y}^{2}}}$ ------- (2) and
$z=\sqrt{2-{{x}^{2}}-{{y}^{2}}}$ ---------- (3)
Now, we will solve equation (2), which is
$z=\sqrt{{{x}^{2}}+{{y}^{2}}}$
So, we will square both sides in the above equation, we get
$\Rightarrow {{z}^{2}}={{\left( \sqrt{{{x}^{2}}+{{y}^{2}}} \right)}^{2}}$
On further solving the above problem, we get
$\Rightarrow {{z}^{2}}={{x}^{2}}+{{y}^{2}}$ --------- (4)
Now, we will solve equation (3), which is
$z=\sqrt{2-{{x}^{2}}-{{y}^{2}}}$
Now, we will take the square on both sides in the above equation, we get
$\Rightarrow {{z}^{2}}={{\left( \sqrt{2-{{x}^{2}}-{{y}^{2}}} \right)}^{2}}$
On further solving the above equation, we get
$\Rightarrow {{z}^{2}}=2-{{x}^{2}}-{{y}^{2}}$
Now, we will add ${{x}^{2}}+{{y}^{2}}$ in the above equation, we get
$\Rightarrow {{z}^{2}}+{{x}^{2}}+{{y}^{2}}=2-{{x}^{2}}-{{y}^{2}}+{{x}^{2}}+{{y}^{2}}$
As we know, the same terms with opposite signs cancel out each other, thus we get
$\Rightarrow {{z}^{2}}+{{x}^{2}}+{{y}^{2}}=2$
Therefore, we get
$\Rightarrow {{x}^{2}}+{{y}^{2}}+{{z}^{2}}=2$ ---------- (5)
Thus, we know that the above equation is an equation of a sphere, it implies the radius is equal to $\sqrt{2}$
Thus, the value of $\rho =\sqrt{2}$ --------- (6)
Therefore, the range of z is $\left[ 0,\sqrt{2} \right]$ -------- (7)
Now, we will solve the equation (4) and (5), to get the value of z, that is we will substitute the value of equation (4) in equation (5), we get
$\Rightarrow {{z}^{2}}+{{z}^{2}}=2$
$\Rightarrow 2{{z}^{2}}=2$
Now, we will divide 2 on both sides in the above equation, we get
$\Rightarrow \dfrac{2}{2}{{z}^{2}}=\dfrac{2}{2}$
On further solving, we get
$\Rightarrow {{z}^{2}}=1$
On taking the square root on both sides in the above equation, we get
$\Rightarrow \sqrt{{{z}^{2}}}=\sqrt{1}$
Therefore, we get
$\Rightarrow z=\pm 1$
Therefore, the value of z is equal to 1 -------- (8)
Also, we know the conversion of z is $\rho \cos \phi $ , therefore we get
$z=\rho \cos \phi $
Thus, we will substitute the value of equation (6) and (8) in the above equation, that is
$\Rightarrow 1=\sqrt{2}\cos \phi $
Now, we will divide $\sqrt{2}$ on both sides in the above equation, we get
$\Rightarrow \dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{\sqrt{2}}\cos \phi $
Therefore, we get
$\Rightarrow \dfrac{1}{\sqrt{2}}=\cos \phi $
Also, we know that cos function is $\dfrac{1}{\sqrt{2}}$ when the angle is equal to $\dfrac{\pi }{4}$ .
Hence $\rho =\dfrac{\pi }{4}$
Thus, the range of $\rho $ is $\left[ 0,\dfrac{\pi }{4} \right]$ ------- (9)
Now, we will find the value of $\theta $ , that is we know that the unit circle is imposed on the xy-plane where the positive side of x-axis is equal to 0. Also, the complete circle has the range equal to $[0,2\pi ]$ . But in this case, the solid given to us is the one-fourth of the circle.
Therefore, the range of $\theta $ is equal to $\left[ 0,\dfrac{\pi }{2} \right]$ ------- (10)
Thus, from equation (7), (9), and (10), we get the new limits for our integration, that is
\[\Rightarrow \int_{0}^{\dfrac{\pi }{4}}{\int_{0}^{\dfrac{\pi }{2}}{\int_{0}^{\sqrt{2}}{{}}}}\] --------- (11)
Now we have to convert the integrand from rectangular to cylindrical coordinates, that is
$\begin{align}
& x=\rho \sin \phi \cos \theta \\
& y=\rho \sin \phi \sin \theta \\
& dzdydx={{\rho }^{2}}\sin \phi \left( d\rho \right)\left( d\theta \right)\left( d\phi \right) \\
\end{align}$ ----------- (12)
Thus, now we will combine the value of equation (11) and (12), to get the required value of the integral into the cylindrical and spherical coordinates, we get
\[\Rightarrow \int_{0}^{\dfrac{\pi }{4}}{\int_{0}^{\dfrac{\pi }{2}}{\int_{0}^{\sqrt{2}}{\left( \rho \sin \phi \cos \theta \right)\left( \rho \sin \phi \sin \theta \right){{\rho }^{2}}\sin \phi \left( d\rho \right)\left( d\theta \right)\left( d\phi \right)}}}\]
On further simplification, we get
\[\Rightarrow \int_{0}^{\dfrac{\pi }{4}}{\int_{0}^{\dfrac{\pi }{2}}{\int_{0}^{\sqrt{2}}{{{\rho }^{4}}{{\sin }^{3}}\phi \cos \theta \sin \theta \left( d\rho \right)\left( d\theta \right)\left( d\phi \right)}}}\]
Therefore, for the integral $\int_{0}^{1}{\int_{0}^{\sqrt{1-{{x}^{2}}}}{\int_{\sqrt{{{x}^{2}}+{{y}^{2}}}}^{\sqrt{2-{{x}^{2}}-{{y}^{2}}}}{xydzdydx}}}$ , its values in cylindrical or spherical coordinates is \[\int_{0}^{\dfrac{\pi }{4}}{\int_{0}^{\dfrac{\pi }{2}}{\int_{0}^{\sqrt{2}}{{{\rho }^{4}}{{\sin }^{3}}\phi \cos \theta \sin \theta \left( d\rho \right)\left( d\theta \right)\left( d\phi \right)}}}\] .
Note: While solving this problem, do mention all the steps properly to avoid calculations. Always remember that z changes to $\rho $ , y changes to $\theta $ and x changes to $\phi $ . Also, do not forget to mention changing the xy given in the integrand in the cylindrical coordinates.
Complete step-by-step solution:
According to the problem, we have to find the cylindrical or spherical coordinates from a definite integral.
Thus, we will apply the polar form to get the solution.
The integral given to us is $\int_{0}^{1}{\int_{0}^{\sqrt{1-{{x}^{2}}}}{\int_{\sqrt{{{x}^{2}}+{{y}^{2}}}}^{\sqrt{2-{{x}^{2}}-{{y}^{2}}}}{xydzdydx}}}$ --------- (1)
So, from equation (1), we get the value of the limit equals to
$\sqrt{{{x}^{2}}+{{y}^{2}}}\le z\le \sqrt{2-{{x}^{2}}-{{y}^{2}}}$
$0\le y\le \sqrt{1-{{x}^{2}}}$
$0\le x\le 1$
Now, we will first solve the z-domain, that is
$z=\sqrt{{{x}^{2}}+{{y}^{2}}}$ ------- (2) and
$z=\sqrt{2-{{x}^{2}}-{{y}^{2}}}$ ---------- (3)
Now, we will solve equation (2), which is
$z=\sqrt{{{x}^{2}}+{{y}^{2}}}$
So, we will square both sides in the above equation, we get
$\Rightarrow {{z}^{2}}={{\left( \sqrt{{{x}^{2}}+{{y}^{2}}} \right)}^{2}}$
On further solving the above problem, we get
$\Rightarrow {{z}^{2}}={{x}^{2}}+{{y}^{2}}$ --------- (4)
Now, we will solve equation (3), which is
$z=\sqrt{2-{{x}^{2}}-{{y}^{2}}}$
Now, we will take the square on both sides in the above equation, we get
$\Rightarrow {{z}^{2}}={{\left( \sqrt{2-{{x}^{2}}-{{y}^{2}}} \right)}^{2}}$
On further solving the above equation, we get
$\Rightarrow {{z}^{2}}=2-{{x}^{2}}-{{y}^{2}}$
Now, we will add ${{x}^{2}}+{{y}^{2}}$ in the above equation, we get
$\Rightarrow {{z}^{2}}+{{x}^{2}}+{{y}^{2}}=2-{{x}^{2}}-{{y}^{2}}+{{x}^{2}}+{{y}^{2}}$
As we know, the same terms with opposite signs cancel out each other, thus we get
$\Rightarrow {{z}^{2}}+{{x}^{2}}+{{y}^{2}}=2$
Therefore, we get
$\Rightarrow {{x}^{2}}+{{y}^{2}}+{{z}^{2}}=2$ ---------- (5)
Thus, we know that the above equation is an equation of a sphere, it implies the radius is equal to $\sqrt{2}$
Thus, the value of $\rho =\sqrt{2}$ --------- (6)
Therefore, the range of z is $\left[ 0,\sqrt{2} \right]$ -------- (7)
Now, we will solve the equation (4) and (5), to get the value of z, that is we will substitute the value of equation (4) in equation (5), we get
$\Rightarrow {{z}^{2}}+{{z}^{2}}=2$
$\Rightarrow 2{{z}^{2}}=2$
Now, we will divide 2 on both sides in the above equation, we get
$\Rightarrow \dfrac{2}{2}{{z}^{2}}=\dfrac{2}{2}$
On further solving, we get
$\Rightarrow {{z}^{2}}=1$
On taking the square root on both sides in the above equation, we get
$\Rightarrow \sqrt{{{z}^{2}}}=\sqrt{1}$
Therefore, we get
$\Rightarrow z=\pm 1$
Therefore, the value of z is equal to 1 -------- (8)
Also, we know the conversion of z is $\rho \cos \phi $ , therefore we get
$z=\rho \cos \phi $
Thus, we will substitute the value of equation (6) and (8) in the above equation, that is
$\Rightarrow 1=\sqrt{2}\cos \phi $
Now, we will divide $\sqrt{2}$ on both sides in the above equation, we get
$\Rightarrow \dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{\sqrt{2}}\cos \phi $
Therefore, we get
$\Rightarrow \dfrac{1}{\sqrt{2}}=\cos \phi $
Also, we know that cos function is $\dfrac{1}{\sqrt{2}}$ when the angle is equal to $\dfrac{\pi }{4}$ .
Hence $\rho =\dfrac{\pi }{4}$
Thus, the range of $\rho $ is $\left[ 0,\dfrac{\pi }{4} \right]$ ------- (9)
Now, we will find the value of $\theta $ , that is we know that the unit circle is imposed on the xy-plane where the positive side of x-axis is equal to 0. Also, the complete circle has the range equal to $[0,2\pi ]$ . But in this case, the solid given to us is the one-fourth of the circle.
Therefore, the range of $\theta $ is equal to $\left[ 0,\dfrac{\pi }{2} \right]$ ------- (10)
Thus, from equation (7), (9), and (10), we get the new limits for our integration, that is
\[\Rightarrow \int_{0}^{\dfrac{\pi }{4}}{\int_{0}^{\dfrac{\pi }{2}}{\int_{0}^{\sqrt{2}}{{}}}}\] --------- (11)
Now we have to convert the integrand from rectangular to cylindrical coordinates, that is
$\begin{align}
& x=\rho \sin \phi \cos \theta \\
& y=\rho \sin \phi \sin \theta \\
& dzdydx={{\rho }^{2}}\sin \phi \left( d\rho \right)\left( d\theta \right)\left( d\phi \right) \\
\end{align}$ ----------- (12)
Thus, now we will combine the value of equation (11) and (12), to get the required value of the integral into the cylindrical and spherical coordinates, we get
\[\Rightarrow \int_{0}^{\dfrac{\pi }{4}}{\int_{0}^{\dfrac{\pi }{2}}{\int_{0}^{\sqrt{2}}{\left( \rho \sin \phi \cos \theta \right)\left( \rho \sin \phi \sin \theta \right){{\rho }^{2}}\sin \phi \left( d\rho \right)\left( d\theta \right)\left( d\phi \right)}}}\]
On further simplification, we get
\[\Rightarrow \int_{0}^{\dfrac{\pi }{4}}{\int_{0}^{\dfrac{\pi }{2}}{\int_{0}^{\sqrt{2}}{{{\rho }^{4}}{{\sin }^{3}}\phi \cos \theta \sin \theta \left( d\rho \right)\left( d\theta \right)\left( d\phi \right)}}}\]
Therefore, for the integral $\int_{0}^{1}{\int_{0}^{\sqrt{1-{{x}^{2}}}}{\int_{\sqrt{{{x}^{2}}+{{y}^{2}}}}^{\sqrt{2-{{x}^{2}}-{{y}^{2}}}}{xydzdydx}}}$ , its values in cylindrical or spherical coordinates is \[\int_{0}^{\dfrac{\pi }{4}}{\int_{0}^{\dfrac{\pi }{2}}{\int_{0}^{\sqrt{2}}{{{\rho }^{4}}{{\sin }^{3}}\phi \cos \theta \sin \theta \left( d\rho \right)\left( d\theta \right)\left( d\phi \right)}}}\] .
Note: While solving this problem, do mention all the steps properly to avoid calculations. Always remember that z changes to $\rho $ , y changes to $\theta $ and x changes to $\phi $ . Also, do not forget to mention changing the xy given in the integrand in the cylindrical coordinates.
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