evaluate triple integral (x^2+2z)dxdydz where T is the region bounded by the pla

Question

Explanation / Answer

puting your values The z bounds are immediate: z = 0 to z = 1 - (x+y)/2. Projecting E onto the xy-plane yields the region between x + y = 2, x = 0, y = 0. ==> y = 0 to y= 2 - x for x in [0, 2]. So, ??? 2y dV = ?(x = 0 to 2) ?(y = 0 to 2 - x) ?(z = 0 to 2 - (x+y)/2) 2y dz dy dx. Evaluating this yields ?(x = 0 to 2) ?(y = 0 to 2 - x) 2y[2 - (x+y)/2] dy dx = ?(x = 0 to 2) ?(y = 0 to 2 - x) (4y - xy + y^2) dy dx = ?(x = 0 to 2) [2y^2 - xy^2/2 + y^3/3] {for y = 0 to 2 - x} dx = ?(x = 0 to 2) [(-5/6) x^3 + 6x^2 - 14x + 32/3] dx = [(-5/24) x^4 + 2x^3 - 7x^2 + 32x/3] {for x = 0 to 2} = 6.

Related Questions

Navigate

• Browse (All)
• Browse E
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.