- #1

Adesh

- 735

- 191

- Homework Statement
- Find the volume under the surface z = \sqrt{1-x^2} and above the triangle formed y = x , x=1 , and the x-axis.

- Relevant Equations
- Volume = double integral of the surface function.

I want to know that how can

Any help would be appreciated. Thank you.

**z=$$ \sqrt{1-x^2}$$**ever represent a surface? It graphs a curve in the x-z plane and the triangle lies in x-y plane so how can they contain a volume, they are orthogonal to each other. I have attached awn image which is drawn GeoGebra for the function**z=$$\sqrt{1-x^2}$$.**My question is if we can write the equation**z = $$\sqrt{1-x^2}$$**as $$\sqrt{1-x^2} - 0.y - z $$. Then how can it traverse in the y-direction, it's coordinate always have to be zero.Any help would be appreciated. Thank you.