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To evaluate the triple integral ∭Dy2dV using Cartesian coordinates, where D is the tetrahedron in the first octant bounded by the coordinate planes and the plane 2x+3y+z=6, we will integrate with respect to x, y, and z over the region D.Given:D is the tetrahedron in the first octant bounded by the coordinate planes and the plane 2x+3y+z=6.dV=dzdydxTo determine the limits of integration, we need to find the intersection points between the plane 2x+3y+z=6 and the coordinate planes.1. Intersection with the x-axis:Setting y=z=0 in the plane equation, we get:2x+3(0)+(0)=62x=6x=32. Intersection with the y-axis:Setting x=z=0 in the plane equation, we get:2(0)+3y+(0)=63y=6y=23. Intersection with the z-axis:Setting x=y=0 in the plane equation, we get:2(0)+3(0)+z=6z=6Based on these intersection points, the limits of integration for x, y, and z are as follows:x:0→3y:0→2z:0→6−2x−3yNow, we can evaluate the integral:∭Dy2dV=∫03∫02∫06−2x−3yy2dzdydxWe will perform the integration in the following order: dz, then dy, and finally dx.Integrating with respect to z first, we have:∫06−2x−3yy2dz=y2·(6−2x−3y)Next, integrating with respect to y, we get:∫02y2·(6−2x−3y)dy=y33·(6−2x−3y)|02Simplifying further, we have:13(2x−3)(2x+1)Finally, integrating with respect to x, we obtain:∫0313(2x−3)(2x+1)dx=13·16(2x−3)2(2x+1)|03Evaluating the limits, we get:13·16(2(3)−3)2(2(3)+1)−13·16(2(0)−3)2(2(0)+1)Simplifying the expression, we find:79Therefore, the value of the given triple integral is 79.

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