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A4-format til udskrift. - Aarhus Universitet

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104 II. INTEGRATION<br />

4.16. Integral over en cirkelring ☞ [S] 12.4 Double integrals in polar . . .<br />

Eksempel 1 - fortsat<br />

(3x + 4y<br />

R<br />

2 π 2<br />

)dA = (3r cos θ + 4r<br />

0 1<br />

2 sin 2 θ)rdr dθ<br />

π 3 4 2 r=2<br />

= r cos θ + r sin θ<br />

r=1<br />

0<br />

dθ<br />

π<br />

= (7cos θ + 15sin<br />

0<br />

2 θ)dθ<br />

π<br />

= (7cos θ +<br />

0<br />

15<br />

(1 − cos 2θ))dθ<br />

2<br />

<br />

= 7sin θ + 15<br />

θ=π (2θ − sin 2θ)<br />

4 θ=0<br />

= 15<br />

2 π<br />

4.17. Integral over en cirkelskive ☞ [S] 12.4 Double integrals in polar . . .<br />

Eksempel 2<br />

Toppen af et æg<br />

{(x,y,z)|x 2 + y 2 ≤ 1, 0 ≤ z ≤ 1 − x 2 − y 2 }<br />

beskrives i "cylinder" koordinater ved<br />

Det er<br />

x = r cos θ, y = r sinθ<br />

{(r,θ,z)|0 ≤ r ≤ 1,0 ≤ z ≤ 1 − r 2 }<br />

4.18. Top ☞ [S] 12.4 Double integrals in polar coordinates<br />

Eksempel 2 - figur<br />

x<br />

z<br />

{(r,θ,z)|0 ≤ r ≤ 1,0 ≤ z ≤ 1 − r 2 }<br />

4.19. Integral over en cirkelskive ☞ [S] 12.4 Double integrals in polar . . .<br />

Eksempel 2 - fortsat<br />

Volumenet er et integral af funktionen<br />

på cirkelskiven<br />

f(x,y) = 1 − x 2 − y 2<br />

R = {(x,y)|x 2 + y 2 ≤ 1}<br />

1<br />

y

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