EP 1) Let D be the unit disc (the region inside the unit circle). Let L and R be the right half of the unit disc respectively, and let T and B be the top andbottom of the unit disc. Decide - without calculation - whether each of the following integrals is positive, negative, or zero.a) S SR 3xd A.d) S SB (y - y³) dA.g) ff cos x d.A.b) f ₁ 3xd A.e) SSD (y – y³) dA.h) S ST-ye¹ d.A.c) S SB 3xdA.f) SSD sin x dAi) f S x cos y dA.B

Note: As per the guidelines, we are supposed to answer only 3 subparts. Please repost the other subparts.D is the region inside the unit circle centered at the origin.L is left half of DR is the right half of D.T is the Top of DB is the bottom of D.(a) Explanation: Since R is the right half of D and region R is symmetrical about the x-axis (the integrand is non-negative in region R because x is non-negative in R).(b) Explanation: Since L is the left half of D and region R is symmetrical about the x-axis (the integrand is negative in region L because x is negative in R).(c) Explanation: Region B is symmetric with respect to the y-axis. At points that are mirror images of each other, relative to the y-axis, the integrand, 3x, takes values that have the same absolute value but opposite signs. The integral over the right half of B will be positive, the integral over the left half of B will be negative, and these two integrals will cancel each other out, so the integral over B will be zero.(d) The region of integration is given by the bottom of the unit circle centered at the origin. In this region, the y coordinates are negative, and the integral we get is also in the form of a function of y, so this integral is negative.(e) The region D is symmetric with respect to the x-axis. At points that are mirror images of each other, one above and one below the x-axis, the integrand will take values that have the same absolute value but opposite signs. The integral over the top half of D will be positive, the integral over the bottom half of D will be negative, and these two integrals will exactly cancel each other out, so the integral over D will be zero.a. Positiveb. negativec. zerod. negativee. zero