2²+42-1 ≤2≤0 p²-1 ≤250 ZE [p² -1,0), PECO, 1] о 4j do j pdp 1 G x + 6 p² sz de p²-1 dz 12π √ pdp[²² + P²²]²²-1 0-(p² - 1)² + p² (p²-1)] 0 - (P" +1-2 p² + P " - P = ²) 4 2 1 x 2 ॰-ངྒཾ,- ) +1-2p²+2p4-2p² 2 6π Sp1-04-1+2p²-284+2p²) dp 1 bπ J-p³- p+2p³ -2p³ +2p³ dp = -3π + гри - + 1 The flux of the vector field F(x, y, z) = (y − z² + 2x³, −z +x²+2y³, x − y²+2z³) out- going from the set {(x, y, z) = R³ : x² + y² - 1 ≤ x ≤ 0} is (A) 20 (B) 0 (C) -π Зп (D) 2

Elementary Linear Algebra (MindTap Course List)
8th Edition
ISBN:9781305658004
Author:Ron Larson
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Chapter5: Inner Product Spaces
Section5.CR: Review Exercises
Problem 47CR: Find an orthonormal basis for the subspace of Euclidean 3 space below. W={(x1,x2,x3):x1+x2+x3=0}
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i attached the question 
and the way i solved it, i believe i made an error, could you point it out for me because the correct answer is 3pi/2
correct answer is D, please see both attached photos

2²+42-1 ≤2≤0
p²-1 ≤250
ZE [p² -1,0), PECO, 1]
о
4j do j pdp 1 G x + 6 p² sz
de
p²-1
dz
12π √ pdp[²² + P²²]²²-1
0-(p² - 1)² + p² (p²-1)]
0 - (P" +1-2 p² + P " - P = ²)
4
2
1 x 2
॰-ངྒཾ,- )
+1-2p²+2p4-2p²
2
6π Sp1-04-1+2p²-284+2p²) dp
1
bπ J-p³- p+2p³ -2p³ +2p³ dp
=
-3π
+
гри
-
+
Transcribed Image Text:2²+42-1 ≤2≤0 p²-1 ≤250 ZE [p² -1,0), PECO, 1] о 4j do j pdp 1 G x + 6 p² sz de p²-1 dz 12π √ pdp[²² + P²²]²²-1 0-(p² - 1)² + p² (p²-1)] 0 - (P" +1-2 p² + P " - P = ²) 4 2 1 x 2 ॰-ངྒཾ,- ) +1-2p²+2p4-2p² 2 6π Sp1-04-1+2p²-284+2p²) dp 1 bπ J-p³- p+2p³ -2p³ +2p³ dp = -3π + гри - +
1 The flux of the vector field F(x, y, z) = (y − z² + 2x³, −z +x²+2y³, x − y²+2z³) out-
going from the set {(x, y, z) = R³ : x² + y² - 1 ≤ x ≤ 0} is
(A)
20
(B) 0
(C) -π
Зп
(D)
2
Transcribed Image Text:1 The flux of the vector field F(x, y, z) = (y − z² + 2x³, −z +x²+2y³, x − y²+2z³) out- going from the set {(x, y, z) = R³ : x² + y² - 1 ≤ x ≤ 0} is (A) 20 (B) 0 (C) -π Зп (D) 2
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