## Moderator: Juilee

Juilee2.2k
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Location:
Dombivli
Last seen:
1 month, 1 week ago
Joined:
2 years, 11 months ago
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j***********@gmail.com

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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 $x^2 +y^2 = a ... written 10 months ago by Juilee2.2k • updated 6 months ago by Sanket Shingote ♦♦ 250 1 answer 417 views 1 answers ... Using spherical coordinate$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 10 months ago by Juilee2.2k
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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 10 months ago by Juilee2.2k
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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 $V = \int^2_0 \int^{2\pi}_0 \int^{arsin\theta}_0 dz \hspace{0.1cm}r\hspace{0.1cm}dr\hspace{0.1cm}d \theta\\ \hspace{0.2cm} = \int^{2a}_{r=0} \int^{2\pi}_{\theta = 0} [z]^{a-rsin\theta}_0 ... written 10 months ago by Juilee2.2k 1 answer 382 views 1 answers ... Let the equation of the sphere be$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 10 months ago by Juilee2.2k
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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 $V = \int^1_{-1}\int^{2\sqrt{1-x^2}}_{-2\sq ... written 10 months ago by Juilee2.2k 1 answer 510 views 1 answers ... Transforming to spherical coordinate by putting$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 \phiI = ...
written 10 months ago by Juilee2.2k
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... $I= \int^1_{x=0}\int^{1-x}_{y=0} \int^{1-x-y}_{z=0}(x+y+z) dx \hspace{0.1cm}dy \hspace{0.1cm}dz\\ \hspace{0.2cm}= \int^{1}_{x=0}\int_{y=0}^{1-x}\Big[\frac{(x+y+z)^2}{2}\Big]^{1-x-y}_0 dx\hspace{0.1cm}dy\\ \hspace{0.2cm}= \frac{1}{2}\int^1_{x=0}\int^{1-x}_{y=0}[1- (x+y)^2]dx \hspace{0.1cm}dy\\ \hspa ... written 10 months ago by Juilee2.2k 1 answer 289 views 1 answers ...$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 10 months ago by Juilee2.2k
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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 10 months ago by Juilee2.2k

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