1. ## Double Integral

Find the volume of the solid lying under the plane z=4x+3y+5 and above the rectangle R={(x,y) | -1<= x <= 0 , 2 <=y <=3}

$\int_{2}^{3}\int_{-1}^{0}~(4x+3y+5) dxdy$

$=~\int_{2}^{3}~[2x^2+3yx+5x]_{-1}^{0}~dy$

$=~\int_{2}^{3}~(-\frac{3y}{2} - \frac{11}{6})dy$

$= ~ [-\frac{3y^2}{4} - \frac{11y}{6}]_{2}^{3}$

$=~-\frac{28}{3}$

which is wrong. The correct answer is $\frac{21}{2}$

Can someone help me? I have a feeling I might have miss used -1,0 and 2,3 maybe.
I also have another question similar to this I will need help on because I'm not quite sure where to start.

2. There's nothing wrong with your set up, must be in the algebra.

I get 21/2.

3. on the last step I even put it in the calculator and get -28/3. Can't seem to figure out where I have messed up.

4. I don't know what you're doing. I am getting 21/2. I even tried various arrangements and did not arrive at -28/3

5. My other question was I need help on this problem as I don't know where to even start. Thanks

Find volume of the solid bounded by the surface
$z=x\sqrt{9x^2+y}$
and the plane x=1,x=0,y=1,y=0 and z=0.

6. Originally Posted by dukebdx12
$=~\int_{2}^{3}~[2x^2+3yx+5x]_{-1}^{0}~dy$

$=~\int_{2}^{3}~(-\frac{3y}{2} - \frac{11}{6})dy$
Check this. (It's just a matter of arithmetic.)

7. Originally Posted by Krizalid
Check this. (It's just a matter of arithmetic.)
got it now...thankyou