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Let I=02[04x2sin(x2+y2)dy]dxI = \int _ { 0 } ^ { 2 } \left[ \int _ { 0 } ^ { \sqrt { 4 - x ^ { 2 } } } \sin \left( x ^ { 2 } + y ^ { 2 } \right) d y \right] d x

Question 161

Multiple Choice

Let I=02[04x2sin(x2+y2) dy]dxI = \int _ { 0 } ^ { 2 } \left[ \int _ { 0 } ^ { \sqrt { 4 - x ^ { 2 } } } \sin \left( x ^ { 2 } + y ^ { 2 } \right) d y \right] d x Then I in polar form is


A) π2π[02rsin(r2) dr]dθ\int _ { \frac { \pi } { 2 } } ^ { \pi } \left[ \int _ { 0 } ^ { 2 } r \sin \left( r ^ { 2 } \right) d r \right] d \theta
B) 0π[02sin(r2) dr]dθ\int _ { 0 } ^ { \pi } \left[ \int _ { 0 } ^ { 2 } \sin \left( r ^ { 2 } \right) d r \right] d \theta
C) 0π[02rsin(r2) dr]dθ\int _ { 0 } ^ { \pi } \left[ \int _ { 0 } ^ { 2 } r \sin \left( r ^ { 2 } \right) d r \right] d \theta
D) 0π2[02rsin(r2) dr]dθ\int _ { 0 } ^ { \frac { \pi } { 2 } } \left[ \int _ { 0 } ^ { 2 } r \sin \left( r ^ { 2 } \right) d r \right] d \theta
E) 0π2[02sin(r2) dr]dθ\int _ { 0 } ^ { \frac { \pi } { 2 } } \left[ \int _ { 0 } ^ { 2 } \sin \left( r ^ { 2 } \right) d r \right] d \theta

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