a. The boundary of the hemisphere is the circle $x^2+y^2=9$ in the plane $z=0$ , where the curl is $\mathrm{curl}\vec F=2y\,\vec\imath+3\,\vec k$ . Green's theorem applies here, so that

$\displaystyle\iint_S\mathrm{curl}\vec F\cdot\mathrm d\vec S=\int_{\partial S}\vec F\cdot\mathrm d\vec r=3\int_{x^2+y^2=9}\mathrm d\vec r$

which means the value of the line integral is 3 times the area of the circle, or $27\pi$ .

b. The closed sphere has no boundary, so by Stokes' theorem the integral is 0.

7 0

3 years ago

Solving Multi-Step Equations

KengaRu [80]

$1.\\2n+3n+7=-41\\5n=-41-7\\5n=-48\\n=-48:5\\\boxed{n=-9.6}\\------------------\\2.\\2x-5x+6.3=-14.4\\-3x=-14.4-6.3\\-3x=-20.7\\x=(-20.7):(-3)\\\boxed{x=6.9}\\------------------$

$3.\\2z+9.75-7z=5.15\\-5z=5.15-9.75\\-5z=-4.6\\z=(-4.6):(-5)\\\boxed{z=0.92}\\------------------\\4.\\3h-5h+11=17\\-2h=17-11\\-2h=6\\h=6:(-2)\\\boxed{h=-3}\\------------------$

$5.\\2t+8-t=-3\\t=-3-8\\\boxed{t=-11}\\------------------\\6.\\6a-2a=-36\\4a=-36\\a=-36:4\\\boxed{a=-9}\\------------------$

$7.\\3c-8c+7=-18\\-5c=-18-7\\-5c=-25\\c=(-25):(-5)\\\boxed{c=5}\\------------------\\8.\\7g+14-5g=-8\\2g=-8-14\\2g=-22\\g=-22:2\\\boxed{g=-11}\\------------------$

$9.\\2b-6+3b=14\\5b=14+6\\5b=20\\b=20:5\\\boxed{b=4}\\------------------\\10.\\2(a-4)+15=13\\2a-8=13-15\\2a=-2+8\\2a=6\\a=6:2\\\boxed{a=3}\\------------------$

$11.\\7+2(a-3)=-9\\2a-6=-9-7\\2a=-16+6\\2a=-10\\a=-10:2\\\boxed{a=-5}\\------------------\\12.\\13+2(5c-2)=29\\10c-4=29-13\\10c=16+4\\10c=20\\c=20:10\\\boced{c=2}$

8 0

3 years ago

What is the value of f(2) for f(x)=2x5−8x4+3x3−4x2+9x−1?(1 point)

Yanka [14]

Answer:

-39

Step-by-step explanation:

just took the quick check

4 0

3 years ago