how can i solve this volume problem??

[arnil9]

hey guys i hav no idea o how to go about this problem??

the solid under the surface z = x^2*y and above the triangle in the xy-plane with vertice at (1, 0), (2, 1), and (4,0)

 

[tkhunny]

1) Determine that the entire surface is on one side of the x-y plane, at least over the triangular region described.
2) Notice that two poins are on the x-axis
3) Write the equations of the two lines containing the two sides not on the x-axis.
4) This is why one invented the "double integral".
5) Describe the triangular region in your integration limits.
6) Never, ever, ever, conclude that you have "no idea". That just can't be right. If your brain is still in your head, you must have some clue. Don't give up again.

 

[soroban]

Hello, arnil9

Find the volume of the solid under the surface $\displaystyle z\,=\,x^2y$ and
above the triangle in the xy-plane with vertices at $\displaystyle A(1,0),\:B(2, 1),\:C(4,0)$

We know that: \(\displaystyle \:V \;=\;\L\int \int\)$\displaystyle x^2y\,dy\,dx$

And we must find the limits for $\displaystyle y$ and $\displaystyle x.$


Graph the region:

        |
        |       B
        |       *(2,1)
        |      /::\
        |     /:::::\
        |    /::::::::\
      - + - * - - - - - * -
          (1,0)       (4,0)
            A           C

The line through A and B has the equation: $\displaystyle \:y \:=\:x\,-\,1$
The line through B and C has the equation: $\displaystyle \:y\:=\:4\,-\,x$


We must construct two integrals:

\(\displaystyle \L V \;=\;\int^{\;\;\;2}_1\int^{\;\;\;\;\;\;x-1}_0 x^2y\,dy\,dx \:+\,\int^{\;\;\;4}_2\int^{\;\;\;\;\;\;\;4-x}_0 x^2y\,dy\,dx\)

 

[arnil9]

thanks alot guys for your help i think i can tackle it from here on...