Calculating Triple Integrals – Bounded by surfaces – Exercise 4559 Post category:Calculating Triple Integral Post comments:0 Comments Exercise Calculate the integral ∫∫∫Tycos(z+x)dxdydz\int\int\int_T y\cos(z+x) dxdydz∫∫∫Tycos(z+x)dxdydz Where T is bounded by the surfaces y=x,y=0,z=0,x+z=π2y=\sqrt{x},y=0,z=0,x+z=\frac{\pi}{2}y=x,y=0,z=0,x+z=2π Final Answer Show final answer ∫∫∫Tycos(z+x)dxdydz=π216−12\int\int\int_T y\cos(z+x) dxdydz=\frac{{\pi}^2}{16}-\frac{1}{2}∫∫∫Tycos(z+x)dxdydz=16π2−21 Solution Coming soon… Share with Friends Read more articles Previous PostCalculating Triple Integrals – Bounded by surfaces – Exercise 4566 Next PostCalculating Triple Integrals – Fixed integration limits – Exercise 4556 You Might Also Like Calculating Triple Integrals – Fixed integration limits – Exercise 4548 April 6, 2019 Calculating Triple Integrals – Fixed integration limits – Exercise 4556 April 6, 2019 Calculating Triple Integrals – Bounded by surfaces – Exercise 4566 April 6, 2019 Calculating Triple Integrals – Bounded by surfaces – Exercise 4573 April 6, 2019 Leave a Reply Cancel replyCommentEnter your name or username to comment Enter your email address to comment Enter your website URL (optional) Δ
