Question

\[\int\limits_{0}^{1}\int\limits_{0}^{x}\sqrt{1+(9/16)+9y ^{2}}dydx\]

Answer 1

you sure this is typed up correctly?

Answer 2

A surface is defined by z=1+3/4x +(3/2)Y^2. Calculate the area of surface directly above the triangle on the xy plane defined by the vertices (0,0), (0,1) and (1,1)

Answer 3

and then i just used the formula for surface area?? no

Answer 4

?

Answer 5

it sounds plausible, but im thinking it thru

Answer 6

we need to define the length of a curve as we move from the xaxis up and as we move over from the y axis

Answer 7

this is one of the ways we could set this up to move across the region

\[\int_{0}^{1}\int_{x}^{1}\sqrt{1+(\frac{3}{4}x)'^2 +(\frac{3}{2}y^2)'^2}dydx\]
\[\int_{0}^{1}\int_{x}^{1}\sqrt{1+\frac{9}{16} +9y^2}\ dydx\]
\[\int_{0}^{1}\int_{x}^{1}\sqrt{\frac{25}{16} +9y^2}\ dydx\]

Answer 8

it has to be x to 1 for dy? why can't it be 0 to x?

Answer 9

the surface above 0 to x might not be the same as the surface above x to 0