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VARVARA [1.3K]
3 years ago
10

Find the surface area of the surface given by the portion of the paraboloid z=3x2+3y2 that lies inside the cylinder x2+y2=4. (hi

nt: convert to polar coordinates after setting up the integral)
Mathematics
1 answer:
natta225 [31]3 years ago
7 0
Parameterize the part of the paraboloid within the cylinder - I'll call it S - by

\mathbf r(u,v)=\langle x(u,v),y(u,v),z(u,v)\rangle=\left\langle u\cos v,u\sin v,3u^2\right\rangle

with 0\le u\le2 and 0\le v\le2\pi. The region's area is given by the surface integral

\displaystyle\iint_S\mathrm dS=\int_{u=0}^{u=2}\int_{v=0}^{v=2\pi}\|\mathbf r_u\times\mathbf r_v\|\,\mathrm du\,\mathrm dv
=\displaystyle\int_{v=0}^{v=2\pi}\int_{u=0}^{u=2}u\sqrt{1+36u^2}\,\mathrm du\,\mathrm dv
=\displaystyle2\pi\int_{u=0}^{u=2}u\sqrt{1+36u^2}\,\mathrm du

Take w=1+36u^2 so that \mathrm dw=72u\,\mathrm du, and the integral becomes

=\displaystyle\frac{2\pi}{72}\int_{w=1}^{w=145}\sqrt w\,\mathrm dw
=\displaystyle\frac\pi{36}\frac23w^{3/2}\bigg|_{w=1}^{w=145}
=\dfrac\pi{54}(145^{3/2}-1)\approx101.522
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