Explanation Given: The solid is bounded by the graphs of the equations z = x , y = x + 2 , y = x 2 , first octant. Formula used: ∫ d z = z + C ∫ d y = y + C ∫ x n d x = x n + 1 n + 1 + C Calculation: In Figure, the projection of the solid onto the xy- plane gives the shaded region which is bounded between the line y = x + 2 and the parabola y = x 2 , such that the bounds for y will be x 2 ≤ y ≤ x + 2 . In order to obtain the bounds of x , we need to find the common intersection point between the parabola and the line in the first octant. This implies that x 2 = x + 2 x 2 − x − 2 = 0 ( x − 2 ) ( x + 1 ) = 0 which gives the values x = 2 or x = − 1 . However, x = 2 will be the solution as we are in the first octant. Thus, the bounds for x will be 0 ≤ x ≤ 2 . As for the plane given by z = x , the bounds for z will be 0 ≤ z ≤ x . Therefore, the volume of the solid is given by V = ∭ Q d V = ∫ 0 2 ∫ x 2 x + 2 ∫ 0 x d z d y d x For the first integration, keeping x and y as constants and integrating with respect to z, we get V = ∫ 0 2 ∫ x 2 x + 2 ( ∫ 0 x d z ) d y <

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Chapter 14.6, Problem 24E

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To calculate: The volume of the solid bounded by the graphs of the equations z=x, y=x+2, y=x2, first octant, using a triple integral.

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Chapter 14.6, Problem 23E

Chapter 14.6, Problem 25E

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Chapter 14 Solutions

EBK CALCULUS: EARLY TRANSCENDENTAL FUNC

Ch. 14.1 - Evaluate the iterated integral: 0433cosrdrd Ch. 14.1 - Prob. 1E Ch. 14.1 - Evaluate the integral: xx2yxdy Ch. 14.1 - Prob. 3E Ch. 14.1 - Prob. 4E Ch. 14.1 - Evaluate the integral: 04x2x2ydy Ch. 14.1 - Evaluate the integral: x3x(x2+3y2)dy Ch. 14.1 - Evaluate the integral: eyyylnxxdx;y0 Ch. 14.1 - Evaluate the integral: 1y21y2(x2+y2)dx Ch. 14.1 - Evaluate the integral: 0x2yeyxdy

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The volume of the solid bounded by the graphs of the equations z = x , y = x + 2 , y = x 2 , first octant, using a triple integral.

Solution Summary: The author calculates the volume of the given solid using a triple integral.

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To calculate: The volume of the solid bounded by the graphs of the equations z=x, y=x+2, y=x2, first octant, using a triple integral.

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Chapter 14 Solutions