Question #271574

Find the surface integral of the vector field ๐‘ญ(๐‘ฅ,๐‘ฆ,๐‘ง)=(๐‘ฅ,๐‘ฆ,๐‘ง) over the part of the paraboloid ๐‘ง=1โˆ’๐‘ฅ^2โˆ’๐‘ฆ^2 with ๐‘งโ‰ฅ0 and having normal pointing upwards.


Hint: take ๐‘ฅ and ๐‘ฆ as independent parameters.


1
Expert's answer
2021-11-28T18:54:51-0500

The surface S can be represented by:

r(x,y)=xi+yj+(1โˆ’x2โˆ’y2)k, โˆ’1โ‰คxโ‰ค1,โˆ’1โ‰คyโ‰ค1r(x, y) = x i + y j + (1 โˆ’ x^ 2 โˆ’ y ^2 ) k,\ โˆ’1 โ‰ค x โ‰ค 1, โˆ’1 โ‰ค y โ‰ค 1

rx=iโˆ’2xk,ry=jโˆ’2ykr_x = i โˆ’ 2x k , r_y = j โˆ’ 2y k

rxร—ry=โˆฃijk10โˆ’2x01โˆ’2yโˆฃ=2xi+2yj+kr_x ร— r_y=\begin{vmatrix} i & j&k \\ 1 & 0&-2x\\ 0 & 1&-2y \end{vmatrix}=2xi+2yj+k


โˆฌSFโ‹…ndS=โˆฌFโ‹…(rxร—ry)dA=โˆซโˆ’11โˆซโˆ’11(2x2+2y2+1โˆ’x2โˆ’y2)dxdy=\iint_SF\cdot ndS=\iint F\cdot (r_x ร— r_y)dA=\int^1_{-1} \int^1_{-1}(2x^2+2y^2+1 โˆ’ x^ 2 โˆ’ y ^2)dxdy=


=โˆซโˆ’11โˆซโˆ’11(x2+y2+1)dxdy=โˆซโˆ’11(2/3+2y2+2)dy==\int^1_{-1} \int^1_{-1}(x^2+y^2+1 )dxdy=\int^1_{-1} (2/3+2y^2+2)dy=


=4/3+4/3+4=20/3=4/3+4/3+4=20/3

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS