## Moderator: Juilee

Juilee ♦♦

**1.9k**- Reputation:
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- Last seen:
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- Email:
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#### Posts by Juilee

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...
$\text{Volume} = 8 \int\int\int dx \hspace{0.1cm}dx \hspace{0.1cm}dy \hspace{0.1cm}dz\\
\hspace{0.1cm}= 8 \int \int \int_{z=0}^{\sqrt{a^2-x^2}}dx \hspace{0.1cm}dy \hspace{0.1cm}dz\\
\hspace{0.1cm} = 8 \int\int(\sqrt{a^2 – x^2})dx \hspace{0.1cm}dy$
Now in the XY plane we have a circle $x^2 +y^2 = a ...

written 6 months ago by
Juilee ♦♦

**1.9k**• updated 10 weeks ago by Sanket Shingote ♦♦**220**0

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... Using spherical coordinate
$X = arsin \theta cos \phi\\
y = br sin \theta sin \phi\\
z = cr cos \theta\\
dx \hspace{0.1cm}dy \hspace{0.1cm}dz = abc r^2 sin \theta dr \hspace{0.1cm}d\theta \hspace{0.1cm}d \phi\\
V = 8 \int_{\phi = 0}^{\pi/2}\int_{\theta = 0}^{\pi/2}\int_{r = 0}^1 dx \hspace{0.1cm}dy ...

written 6 months ago by
Juilee ♦♦

**1.9k**0

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... If we take projections on the XY plane, the area is bounded by the circle $x^2 + y^2 = 2$, the line y = x and the line x = 0.
We change the coordinate to cylindrical polar by putting x = $rcos \theta, y = rsin \theta$, z = z .
Eqn. of the cylinder becomes $x^2 + y^2 = 2 \Rightarrow r = \sqrt{2}$
t ...

written 6 months ago by
Juilee ♦♦

**1.9k**0

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... z varies from z = 0 to $z = a-y = a – r sin \theta$
$\theta$ varies from 0 to 2$\pi$ and r varies from 0 to 2
$V = \int^2_0 \int^{2\pi}_0 \int^{arsin\theta}_0 dz \hspace{0.1cm}r\hspace{0.1cm}dr\hspace{0.1cm}d \theta\\
\hspace{0.2cm} = \int^{2a}_{r=0} \int^{2\pi}_{\theta = 0} [z]^{a-rsin\theta}_0 ...

written 6 months ago by
Juilee ♦♦

**1.9k**0

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... Let the equation of the sphere be $x^2 + y^2 + z^2 =a^2$. Using cylindrical polar coordinate
$x = rcos \theta, y = rsin \theta, z = z$
we see that in the first octant z varies from z = 0 to $z = \sqrt{a^2 - (x^2+y^2)} = \sqrt{a^2 - r^2}$ r varies from r = b to r = a and $\theta$ varies from $\thet ...

written 6 months ago by
Juilee ♦♦

**1.9k**0

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... The xy plane cut the parabolid in the ellipse $x^2 + \frac{y^2}{4} = 1$
Hence the total volume
$v = \iint_R z \hspace{0.1cm}dx\hspace{0.1cm}dy\\
\hspace{0.1cm}=\iint_R\Big(1 - x^2-\frac{y^2}{4}\Big)dx \hspace{0.1cm}dy$
Where R i the area of the ellipse
$V = \int^1_{-1}\int^{2\sqrt{1-x^2}}_{-2\sq ...

written 6 months ago by
Juilee ♦♦

**1.9k**0

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... Transforming to spherical coordinate by putting $x = r \hspace{0.1cm}sin \theta cos \theta , y=r \hspace{0.1cm}sin \theta \hspace{0.1cm}sin \phi, z=r \hspace{0.1cm}cos \theta$ and $dx\hspace{0.1cm}dy\hspace{0.1cm}dz = r^2sin \theta \hspace{0.1cm}dr \hspace{0.1cm}d \theta \hspace{0.1cm}d \phi$
$I = ...

written 6 months ago by
Juilee ♦♦

**1.9k**0

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...
$I= \int^1_{x=0}\int^{1-x}_{y=0} \int^{1-x-y}_{z=0}(x+y+z) dx \hspace{0.1cm}dy \hspace{0.1cm}dz\\
\hspace{0.2cm}= \int^{1}_{x=0}\int_{y=0}^{1-x}\Big[\frac{(x+y+z)^2}{2}\Big]^{1-x-y}_0 dx\hspace{0.1cm}dy\\
\hspace{0.2cm}= \frac{1}{2}\int^1_{x=0}\int^{1-x}_{y=0}[1- (x+y)^2]dx \hspace{0.1cm}dy\\
\hspa ...

written 6 months ago by
Juilee ♦♦

**1.9k**0

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...
$I = \int_{x=0}^2\int^x_{y=0}\int^{2x+2y}_{z=0}e^{x+y+z}dz\hspace{0.1cm}dy\hspace{0.1cm}dx\\
\hspace{0.1cm}= \int_0^2 \int_0^x e^{x+y}[e^z]_0^{2x+2y} dx \hspace{0.1cm}dy\\
\hspace{0.1cm}= \int_0^2\int_0^x e^{x+y}[e^{2x+2y}-e^0]dx \hspace{0.1cm}dy\\
\hspace{0.1cm}= \int_0^2\int_0^x(e^{3x+3y}-e^{x+y} ...

written 6 months ago by
Juilee ♦♦

**1.9k**0

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... Let $p(r, \theta)$ be any point on the given cardiode. The distance of p from it axis is $y = rsin \theta$. The density at any point is $p = kr^2sin^2 \theta$.
Mass of the lamina $= \int^{\pi}_{\theta = 0}\int_0^{a(1+cos \theta)}Kr^2 sin^2 \theta r \hspace{0.1cm}dr \hspace{0.1cm}d \theta\\
= 2k \in ...

written 6 months ago by
Juilee ♦♦

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