This is because to follow pemdas you do the parenthesis first so you subtract both of the shorter sides. Next since there are two longer sides you divide by two to get the answer. Each longer side is equal to 36 inches.

Hope this helps

6 0

2 years ago

Given that curl F = 2yi – 2zj + 3k, find the surface integral of the normal component of curl F (not F) over (a) the open hemisp

Dimas [21]

Use Stokes' theorem for both parts, which equates the surface integral of the curl to the line integral along the surface's boundary.

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

2 years ago