Volume of solid with double integrals

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SUMMARY

The volume of the solid bounded by the equations z=x, y=x, x+y=2, and z=0 is calculated using double integrals. The domain for x is established as 0 < x < 1, and for y, it is defined as x < y < 2-x. The integral to compute the volume is set up as ∬ x dy dx. Upon integrating first with respect to y, the result simplifies to x^2/2, yielding a final volume of 1/2 after evaluating the bounds. This confirms that the integration process is correct and the final answer is indeed 1/2.

PREREQUISITES
  • Understanding of double integrals in calculus
  • Familiarity with the concept of bounded regions in three-dimensional space
  • Knowledge of setting up limits of integration
  • Basic proficiency in evaluating integrals
NEXT STEPS
  • Study the method of finding volumes using double integrals in multivariable calculus
  • Learn about the geometric interpretation of double integrals
  • Explore the application of Jacobians in changing variables for double integrals
  • Practice problems involving volume calculations of solids bounded by multiple surfaces
USEFUL FOR

Students studying calculus, particularly those focusing on multivariable calculus and applications of double integrals in finding volumes of solids.

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Homework Statement



find volume of solid bounded by z=x, y=x, x+y=2 and z=0


The Attempt at a Solution



first need to find domain.

for x bounds, when y=0, x=0, when y = x, x+x=2 so x=1 therefore 0 < x < 1
for y bounds, x < y < 2-x

now I am trying to work out what i integrate over. usually its the plane z, so if z=x then the integral is SS x.dxdy

is that right?
 
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now i know that's the right integral. but when i integrate first with respect to y, there is no y.

so just integrate x with respect to x which is x^2/2 then sub in 1 and the final answer is 1/2?

but I am not sure if its that simple?
 

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