Set Up an Iterated Integral For $\int _ { W } f ( x , y , z ) d V$

question 24

Multiple Choice

Set up an iterated integral for $\int _ { W } f ( x , y , z ) d V$ , where W is the solid region bounded below by the rectangle 0 $\le$ x $\le$ 3, 0 $\le$ y $\le$ 1 and above by the surface $z ^ { 2 } + y ^ { 2 } = 1$

• A) \(\int _ { W } f ( x , y , z ) d V = \int _ { 0 } ^ { 3 } \int _ { 0 } ^ { 1 } \int _ { 0 } ^ { \sqrt { 1 - y ^ { 2 } } } f ( x , y , z ) d z c x \alpha d y\)
• B) \(\int _ { W } f ( x , y , z ) d V = \int _ { 0 } ^ { 3 } \int _ { 0 } ^ { 1 } \int _ { 0 } ^ { \sqrt { 1 - y ^ { 2 } } } f ( x , y , z ) d y d z d x\)
• C) \(\int _ { W } f ( x , y , z ) d V = \int _ { 0 } ^ { 3 } \int _ { 0 } ^ { 1 } \int _ { 0 } ^ { \sqrt { 1 - y ^ { 2 } } } f ( x , y , z ) d z d y d x\)
• D) \(\int _ { W } f ( x , y , z ) d V = \int _ { 0 } ^ { 3 } \int _ { 0 } ^ { 1 } \int _ { - \sqrt { 1 - y ^ { 2 } } } ^ { \sqrt { 1 - y ^ { 2 } } } f ( x , y , z ) d z d y d x\)
• E) \(\int _ { W } f ( x , y , z ) d V = \int _ { 0 } ^ { 3 } \int _ { 0 } ^ { 1 } \int _ { - \sqrt { 1 - y ^ { 2 } } } ^ { \sqrt { 1 - y ^ { 2 } } } f ( x , y , z ) d y d x d z\)

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