Question #120479
Determine the volume of a solid which is below the plane z = 8– 2x –3y and above the region R in the xy plane define by:
0 ≤ x ≤ 2; 0 ≤ y ≤ 1.
1
Expert's answer
2020-06-08T10:28:11-0400

We should calculate the triple integral (see https://tutorial.math.lamar.edu/classes/calcIII/TripleIntegrals.aspx)

V=02dx01dy082x3ydz=02dx01(82x3y)dy=02(8y2xy3y22)01dx=02(6.52x)dx=(6.5xx2)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.


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