We should calculate the triple integral (see https://tutorial.math.lamar.edu/classes/calcIII/TripleIntegrals.aspx)
V=∫02dx∫01dy∫08−2x−3ydz=∫02dx∫01(8−2x−3y) dy=∫02(8y−2xy−3y22)∣01 dx=∫02(6.5−2x) dx=(6.5x−x2)∣02=9.V = \int\limits_0^2dx\int\limits_0^1 dy\int\limits_0^{8-2x-3y}dz = \int\limits_0^2dx\int\limits_0^1 (8-2x-3y)\,dy = \int\limits_0^2\left(8y-2xy-3\dfrac{y^2}{2}\right)\Big|_0^1\,dx = \int\limits_0^2(6.5-2x)\,dx = \left(6.5x- x^2\right)\Big|_0^2 = 9.V=0∫2dx0∫1dy0∫8−2x−3ydz=0∫2dx0∫1(8−2x−3y)dy=0∫2(8y−2xy−32y2)∣∣01dx=0∫2(6.5−2x)dx=(6.5x−x2)∣∣02=9.
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments