Evaluate the triple integral ∭xdV where the region is the solid bounded by the paraboloid x=5y^2+5z^2 and x=5

Answer 1

for x=5 and x= 5y² +5z² equating these two functions gives 1= y² + z² and this gives -1≤y≤1, -√(x/5-y²)≤z≤√(x/5-y²), 5y²≤x≤5 so the integral becomes ∫_-1¹∫_5x^2⁵∫_{-√(x/5-y^2)}^{√(x/5 -y^2)} dzdxdy. This form is quite complicated to solve so use the following form instead letting x=5 and x= 5y² + 5z² so that integral becomes ∫

∫∫∫5y^2+5z^2 ⁵dx dAwhich becomes ƒƒ(5-5y²-5z²)dA now use polar coordinates where y=1cosθ and z=1sinθ.

Now the integral becomes 5∫₀^2π∫₀¹(1-r²)rdrdθ which equals 10π∫₀¹(r-r³)dr= 10π(r²/2 -r⁴/4)₀¹ =10π(1/2 -1/4)which equals 5π/2.