just the answer plz .. for Q 2 just answer part C ? Let D be the region in the x

Question

just the answer plz .. for Q 2 just answer part C ?

Let D be the region in the x,y-plane enclosed by the curves and . Find the volume of the solid with base D that lies under the surface .
Volume = Let D be the region in the x,y-plane enclosed by the curves just the answer plz .. for Q 2 just answer part C ? displaystyle iint limits_R f(x,y) , dA = int_A^B ! ! int_C^D f(x,y) , dy , dx A = -1 B = 2 C = D = 3 R . Find the limits of integration for the following iterated integral. (a) f(x,y) is a continuous function on R is the shaded region in the figure, and z = 3 y^2 . Volume = Suppose x = 2y - 2 . Find the volume of the solid with base D that lies under the surface x = y^2 - 5 and

Explanation / Answer

1.
y ? 5 = 2y ? 2
y ? 2y ? 3 = 0
(y + 1) (y ? 3) = 0
y = ?1, y = 3

For y on interval [?1, 3] (y ? 5) ? (2y ? 2)

Limits:
?1 ? y ? 3
y ? 5 ? x ? 2y ? 2

V = ? [?1 to 3] ? [y?5 to 2y?2] * 3y dx dy
V = ? [?1 to 3] [2y ? 2 ? y + 5] * 3y dy
V = ? [?1 to 3] [6y3 ?6y ? 3y4 + 15y]dy
V = ? [?1 to 3] [6y3 - 3y4 + 9y]dy
V = 3/2 y4 - 3/5 y5 + 3y3..........apply limit from -1 to 3
V = [243/2 - 729/5 + 81 - 3/2 - 3/5 + 3]
V = [204 - 732/5]
V = 288/5 = 57.6

2.

-1 is the correct answer for C

Graph shows that the value of y varies from -1 to 3

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