Exercises — Triple integrals in Cartesian, cylindrical, spherical coordinates

Region KAUNSI hai: radius $3$ ka ek round tube (cylinder) jo do flat lids ke beech baitha hai. $xy$-plane mein round, $z$-axis ke seedha upar — yeh cylindrical coordinates ka textbook case hai.

KYUN cylindrical: $x^2+y^2=9$ ek simple clean statement $r=3$ ban jaata hai. Cartesian mein hum $y=\pm \sqrt{9-x^2}$ (ek square root) se ladenge, jo bina kisi reason ke zyada ugly hai.

Limits:

• $r$ axis $r=0$ se enter karta hai, wall $r=3$ se exit karta hai.
• $\theta$ ek full circle sweep karta hai: $0\to 2\pi$.
• $z$ dono lids ke beech run karta hai: $0\to 4$.

${\int }_0^{2\pi }{\int }_0^3{\int }_0^{4}f\underset{d{V}_{\text{cyl}}}{\underbrace{rdzdrd\theta }}.$

KAUNSI: radius $5$ ki ek ball, har direction mein round — spherical coordinates ki pehchaan.

KYUN: $x^2+y^2+z^2\le 25$ simply $\rho \le 5$ ban jaata hai. Har boundary ek constant hai.

Limits: $\rho :0\to 5$, $\varphi :0\to \pi$ (top pole se bottom pole tak), $\theta :0\to 2\pi$ (poora around). ${\int }_0^{2\pi }{\int }_0^{\pi }{\int }_0^{5}f{\rho }^2\sin \varphi d\rho d\varphi d\theta .$

Region kaisi dikhti hai — neeche figure dekho. Axis se fixed distance $r$ ke liye, solid slanted cone surface $z=r$ (bottom) aur flat cap $z=2$ (top) ke beech rehta hai. Cone cap se milta hai jahan $r=z=2$, isliye $r$ range karta hai $0\to 2$.

Set up $f=1$ (volume) aur $d{V}_{\text{cyl}}=rdzdrd\theta$ ke saath: $V={\int }_0^{2\pi }{\int }_0^2{\int }_r^{2}rdzdrd\theta .$

Inner ($z$): integrand $r$, $z$ mein constant hai, isliye ${\int }_r^{2}rdz=r(2-r).$ KYUN: ek constant ko ek length par integrate karna simply length $(2-r)$ se multiply kar deta hai.

Middle ($r$): ${\int }_0^2(2r-r^2)dr=[r^2-\frac{r^3}{3}{]}_0^2=4-\frac{8}{3}=\frac{4}{3}.$

Outer ($\theta$): ${\int }_0^{2\pi }d\theta =2\pi$ se multiply karo. $V=2\pi \cdot \frac{4}{3}=\boxed{\frac{8\pi }{3}}.$ Sanity check: base radius $R$ aur height $H$ wale cone ka volume $\frac{1}{3}\pi R^{2}H$ hota hai. Yahan $R=2,H=2$ deta hai $\frac{1}{3}\pi \cdot 4\cdot 2=\frac{8\pi }{3}$. ✓

KYUN cylindrical: integrand $x^2+y^2$ exactly $r^2$ hai — rotational symmetry region aur integrand dono mein hai.

Rewrite: $(x^2+y^2)dV=r^2\cdot rdzdrd\theta =r^{3}dzdrd\theta$. Jacobian $r$ mat bhoolo: yeh $r^2$ se multiply karke $r^3$ deta hai.

${\int }_0^{2\pi }{\int }_0^1{\int }_0^3{r}^{3}dzdrd\theta .$ Saare limits constant hain aur integrand factor karta hai, isliye teen 1D integrals multiply karo: $\underset{2\pi }{\underbrace{{\int }_0^{2\pi }d\theta }}\underset{1/4}{\underbrace{{\int }_0^1{r}^{3}dr}}\underset{3}{\underbrace{{\int }_0^{3}dz}}=2\pi \cdot \frac{1}{4}\cdot 3=\boxed{\frac{3\pi }{2}}.$

Region kaisi hai — figure dekho. Ek bowl (paraboloid) jiske upar $z=4$ pe flat lid hai. Fixed $(r,\theta )$ ke liye ek vertical arrow bowl ke $z=r^2$ pe enter karta hai aur lid pe $z=4$ pe exit karta hai.

JAHAN shadow khatam hota hai: bowl lid se milta hai jab $r^2=4$, yaani $r=2$. Isliye $xy$-shadow disk $r\le 2$ hai.

KYUN cylindrical: $z=x^2+y^2$ hai $z=r^2$ (clean), aur shadow ek disk hai.

$V={\int }_0^{2\pi }{\int }_0^2{\int }_{r^2}^{4}rdzdrd\theta .$ Inner ($z$): ${\int }_{r^2}^{4}rdz=r(4-r^2)$. Middle ($r$): ${\int }_0^2(4r-r^3)dr=[2r^2-\frac{r^4}{4}{]}_0^2=8-4=4.$ Outer ($\theta$): $\times 2\pi$. $V=2\pi \cdot 4=\boxed{8\pi }.$

KAUNSA: ek "ice-cream-cone shell" — ek ball ka wedge jo do cones ke beech trap hai (polar angle $\varphi$ se measure karke) aur radius $2$ tak.

KYUN spherical: boundaries hain $\rho =2$, $\varphi =\pi /6$, $\varphi =\pi /3$ — saari constant coordinate surfaces. Perfect.

Integrand rewrite karo: $z=\rho \cos \varphi$, aur $dV={\rho }^2\sin \varphi d\rho d\varphi d\theta$, isliye $zdV=(\rho \cos \varphi )({\rho }^2\sin \varphi )d\rho d\varphi d\theta ={\rho }^3\sin \varphi \cos \varphi d\rho d\varphi d\theta .$ $I={\int }_0^{2\pi }{\int }_{\pi /6}^{\pi /3}{\int }_0^2{\rho }^3\sin \varphi \cos \varphi d\rho d\varphi d\theta .$ $\rho$-integral: ${\int }_0^2{\rho }^{3}d\rho =\frac{2^4}{4}=4$. $\varphi$-integral: $\sin \varphi \cos \varphi =\frac{1}{2}\sin 2\varphi$ use karo: ${\int }_{\pi /6}^{\pi /3}\frac{1}{2}\sin 2\varphi d\varphi =\frac{1}{2}[-\frac{\cos 2\varphi }{2}{]}_{\pi /6}^{\pi /3}=-\frac{1}{4}(\cos \frac{2\pi }{3}-\cos \frac{\pi }{3}).$ Ab $\cos \frac{2\pi }{3}=-\frac{1}{2}$ aur $\cos \frac{\pi }{3}=\frac{1}{2}$, isliye bracket hai $-\frac{1}{2}-\frac{1}{2}=-1$, jo deta hai $-\frac{1}{4}\cdot (-1)=\frac{1}{4}$. $\theta$-integral: ${\int }_0^{2\pi }d\theta =2\pi$. $I=2\pi \cdot \frac{1}{4}\cdot 4=\boxed{2\pi }.$

KYUN spherical: ball region ($\rho \le a$) plus constraint $z\ge 0$, $\varphi \in [0,\frac{\pi }{2}]$ ban jaata hai (upper half). Integrand $z=\rho \cos \varphi$ spherical mein clean hai.

Denominator (volume) Exercise 2.1 hai: $V=\frac{8\pi }{3}$.

Numerator ${\iiint }_{E}zdV={\int }_0^{2\pi }{\int }_0^2{\int }_r^{2}zrdzdrd\theta .$ Inner ($z$): ${\int }_r^{2}zrdz=r\cdot [\frac{z^2}{2}{]}_r^2=r\cdot \frac{4-r^2}{2}=2r-\frac{r^3}{2}.$ Middle ($r$): ${\int }_0^2(2r-\frac{r^3}{2})dr=[r^2-\frac{r^4}{8}{]}_0^2=4-2=2.$ Outer ($\theta$): $\times 2\pi$ $\Rightarrow \iiint zdV=4\pi$.

$\bar{z}=\frac{4\pi }{8\pi /3}=4\pi \cdot \frac{3}{8\pi }=\boxed{\frac{3}{2}}.$ Sanity check: cone $z\in [0,2]$ par span karta hai lekin top ke paas mota hai (radius $z$ ke saath badhta hai), isliye balance point midpoint $1$ se upar hota hai. Indeed $\bar{z}=1.5>1$. ✓

KYUN spherical single sane choice hai: $(x^2+y^2+z^2{)}^{3/2}=({\rho }^2{)}^{3/2}={\rho }^3$. Woh exponent ${\rho }^3$ exactly wahi hai jisko Jacobian ka ${\rho }^{2}d\rho$ handle karne ke liye bana hai.

$I={\int }_0^{2\pi }{\int }_0^{\pi }{\int }_0^1{e}^{{\rho }^3}{\rho }^2\sin \varphi d\rho d\varphi d\theta .$ Angular part factor out ho jaata hai: ${\int }_0^{2\pi }d\theta \cdot {\int }_0^{\pi }\sin \varphi d\varphi =2\pi \cdot 2=4\pi .$ Radial part — trick hai substitution $u={\rho }^3$, $du=3{\rho }^{2}d\rho$, isliye ${\rho }^{2}d\rho =\frac{1}{3}du$: ${\int }_0^1{e}^{{\rho }^3}{\rho }^{2}d\rho =\frac{1}{3}{\int }_0^1{e}^{u}du=\frac{1}{3}(e-1).$ KYUN yeh substitution: Jacobian ka leftover ${\rho }^2$ exactly $\frac{1}{3}du$ hai — yeh integral designed tha takeable hone ke liye sirf ${\rho }^2$ factor ki wajah se. Spherical ke Jacobian ke bina koi clean $u$-substitution nahi hai. $I=4\pi \cdot \frac{1}{3}(e-1)=\boxed{\frac{4\pi }{3}(e-1)}.$

Region kaisi hai — ek ice-cream cone: ek spherical cap jo $45^{\circ }$ cone par baithi hai. Cone aur sphere dono spherical mein coordinate surfaces hain, isliye use karo.

Cone ke liye $\varphi$ nikalo. Cone $z=\sqrt{x^2+y^2}$ ka matlab $z=r$ hai, yaani $+z$ axis se angle $45^{\circ }=\frac{\pi }{4}$ hai. Spherical se check karo: $z=\rho \cos \varphi$ aur $r=\rho \sin \varphi$, isliye $z=r\Rightarrow \cos \varphi =\sin \varphi \Rightarrow \varphi =\frac{\pi }{4}$. Solid cone ke andar hai (axis ke paas), isliye $\varphi :0\to \frac{\pi }{4}$.

Sphere ke liye $\rho$ nikalo: $x^2+y^2+z^2=8\Rightarrow {\rho }^2=8\Rightarrow \rho =2\sqrt{2}$. Isliye $\rho :0\to 2\sqrt{2}$.

$V={\int }_0^{2\pi }{\int }_0^{\pi /4}{\int }_0^{2\sqrt{2}}{\rho }^2\sin \varphi d\rho d\varphi d\theta .$

• ${\int }_0^{2\sqrt{2}}{\rho }^{2}d\rho =\frac{(2\sqrt{2}{)}^3}{3}=\frac{16\sqrt{2}}{3}$ (kyunki $(2\sqrt{2}{)}^3=8\cdot 2\sqrt{2}=16\sqrt{2}$).
• ${\int }_0^{\pi /4}\sin \varphi d\varphi =[-\cos \varphi {]}_0^{\pi /4}=1-\frac{\sqrt{2}}{2}$.
• ${\int }_0^{2\pi }d\theta =2\pi$.

Sphere-shaped integrand $(x^2+y^2+z^2{)}^p$ ke liye kaunsa $dV$? ::: Spherical, kyunki $(x^2+y^2+z^2{)}^p={\rho }^{2p}$, Jacobian ke ${\rho }^{2}d\rho$ ke saath pair karke ek clean radial integral deta hai. Paraboloid ka shadow radius kahan se aata hai? ::: Paraboloid ko capping plane se intersect karo aur solve karo, jaise $r^2=4\Rightarrow r=2$. Cone ka center of mass uske midpoint se upar kyun hota hai? ::: Solid top ke paas zyada mota (zyada volume) hota hai, isliye balance point upar shift hota hai: $\bar{z}=\frac{3}{2}>1$.