Please answer question EP2. Please give full explanation to the answer. **EP 1)** Find the *mass* of the solid that lies in the first octant below the plane \(2x + 3y + z = 6\), given that the solid's density is given by \(\delta(x, y, z) = x + y\).**EP 2)** Let \( R \) be the solid cone bounded by \( z = \sqrt{x^2 + y^2} \) and \( z = 2 \). Without doing any calculations, decide whether the following integrals are positive, negative, or zero.a) \(\iiint_{R} \sqrt{x^2 + y^2} \, dV \).b) \(\iiint_{R} x \, dV \).c) \(\iiint_{R} (z - 2) \, dV \).d) \(\iiint_{R} xyz \, dV \).

Name: Please answer question EP2. Please give full explanation to the answer
Uploaded: 2024-03-20T07:07:37.000Z
Channel: Bartleby
Description: Please answer question EP2. Please give full explanation to the answer. **EP 1)** Find the *mass* of the solid that lies in the first octant below the plane \(2x + 3y + z = 6\), given that the solid's density is given by \(\delta(x, y, z) = x + y\).*

EP2 Given R is the solid cone, which is bounded by z=x2+y2 and z=2 Figure isSolution for part a: Given ∫∫∫Rx2+y2dV which can be written as ∫∫∫Rx2+y2dV=∫∫∫Rzdxdydz From the figure 0≤z≤2, -z≤y≤z and -z2-y2≤x≤z2-y2 So the above integral convert into ∫∫∫Rx2+y2dV=∫z=02∫y=-zz∫x=-z2-y2z2-y2zdxdydz in the above equation the all upper limit value is greater than the lower limit value, so the integral is positive. Solution for part b: Given ∫∫∫RxdV=∫z=02∫y=-zz∫x=-z2-y2z2-y2xdxdydz In the above integral, value of integral will be zero because the function in the integration is odd and the integration limit is -a to +a type. Solution for part c: ∫∫∫RxdV=∫z=02∫y=-zz∫x=-z2-y2z2-y2(z-2)dxdydz In the above integral, value of integral will be negative because when we do integration we see that factor of higher denominator subtracted by factor of lower denominator. Solution for part d: ∫∫∫RxdV=∫z=02∫y=-zz∫x=-z2-y2z2-y2xyzdxdydz In the above integral, value of integral will be zero because the function in the integration is odd and the integration limit is -a to +a type.