HI THERE!!!!!

The answer is your question is the second choice 59 in^2.

Here is how you can do it:

1. Area of rectangle = length*width = 7*4 = 28 square inches
2. Area of triangle on the top = 0.5*base*height = 0.5*7*6 = 21 square inches
3. Area of triangle on right side = 0.5*base*height = 0.5*4*5 = 10 square inches.

Now you can do the total area:
Add them all →→ 28+21+10 = 59

Hope this helps u out.;)P.S. id really appreciate brainliest thx

~ TRUE BOSS

7 0

2 years ago

Evaluate the surface integral. s x ds, s is the part of the plane 18x + 9y + z = 18 that lies in the first octant.

IceJOKER [234]

In the first octant, the given plane forms a triangle with vertices corresponding to the plane's intercepts along each axis.

$(x,0,0)\implies 18x+9\cdot0+0=18\implies x=1$
$(0,y,0)\implies 18\cdot0+9y+0=18\implies y=2$
$(0,0,z)\implies 18\cdot0+9\cdot0+z=18\implies z=18$

Now that we know the vertices of the surface $\mathcal S$ , we can parameterize it by

$\mathbf s(u,v)=\langle(1-u)(1-v),2u(1-v),18v\rangle$

where $0\le u\le1$ and $0\le v\le1$ . The surface element is

$\mathrm dS=\|\mathbf s_u\times\mathbf s_v\|\,\mathrm du\,\mathrm dv=2\sqrt{406}(1-v)\,\mathrm du\,\mathrm dv$

With respect to our parameterization, we have $x(u,v)=(1-u)(1-v)$ , so the surface integral is

$\displaystyle\iint_{\mathcal S}x\,\mathrm dS=2\sqrt{406}\int_{u=0}^{u=1}\int_{v=0}^{v=1}(1-u)(1-v)^2\,\mathrm dv\,\mathrm du=\frac{\sqrt{406}}3$

5 0

2 years ago