T thebenji New member Joined Sep 2, 2006 Messages 31 Oct 22, 2006 #1 ∫∫R (6x+2y+12)dA where R is the region: [0,1] x [0,3] Here's how I do it: =∫∫R (6x+2y+12)dydx =∫(6x+y²+12y] evaluated from 0 to 3 * dx =∫ (6x+9+36)-(6x+0+0)dx =∫45dx =45x evaluated from 0 to 1 =45-0 =45 What am I doing wrong?
∫∫R (6x+2y+12)dA where R is the region: [0,1] x [0,3] Here's how I do it: =∫∫R (6x+2y+12)dydx =∫(6x+y²+12y] evaluated from 0 to 3 * dx =∫ (6x+9+36)-(6x+0+0)dx =∫45dx =45x evaluated from 0 to 1 =45-0 =45 What am I doing wrong?
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,203 Oct 22, 2006 #2 Your notation is wacky. Is this it: \(\displaystyle \L\\\int_{0}^{1}\int_{0}^{3}(6x+2y+12)dydx\) \(\displaystyle \L\\\int_{0}^{3}(6x+2y+12)dy=|_{0}^{3}(y^{2}+6xy+12y)=18x+45\) \(\displaystyle \L\\\int_{0}^{1}(18x+45)dx=54\)
Your notation is wacky. Is this it: \(\displaystyle \L\\\int_{0}^{1}\int_{0}^{3}(6x+2y+12)dydx\) \(\displaystyle \L\\\int_{0}^{3}(6x+2y+12)dy=|_{0}^{3}(y^{2}+6xy+12y)=18x+45\) \(\displaystyle \L\\\int_{0}^{1}(18x+45)dx=54\)
T thebenji New member Joined Sep 2, 2006 Messages 31 Oct 22, 2006 #3 Oh! I didn't truly treat the 6x as a constant when I integrated...I pulled it out as 6x instead of making it 6xy like any other constant. Thanks!
Oh! I didn't truly treat the 6x as a constant when I integrated...I pulled it out as 6x instead of making it 6xy like any other constant. Thanks!
