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To begin, we divide the region R we will use the outline of the contour plot into 16 equal squares. Another interpretation: If f x is the density at x of a wire and the density is increasing as x increases for x in [a, b], then the center of mass of the wire is to the right of the midpoint of [a, b].
Making use of symmetry, the volume is The integral over S of x 4 y is 0 since this is an odd function of y. Therefore, the Therefore, integral equals 0. Let S ' be the part of S in the first quadrant. We will assume that the cross-section of the river is roughly the shape of an isosceles triangle and that the cross-sectional area is uniform across a slice. Thus, 1 5. Regions A and B are congruent but region B is farther from the origin, so it generates a larger solid than region A generates.
Therefore, the integral is negative. Normal vector to plane is 0, -sin a, cos a. Choose a coordinate system so the center of the sphere is the origin and the axis of the part removed is the z-axis. Problem Set Because the slopes of both roofs are the same, the area of Tm will be the same for both roofs.
Therefore, the area of the roofs will be 2a the same. Thus the surface b. The surface area of a paraboloid and a hyperbolic surface and makes an acute angle with the z-axis.
Then the normal So, the areas depend on the regions. Then obtain that its equation is See note with next Note that this is 0 not valid if we are concerned with values of or moments or mass. See note with previous The moment of inertia with respect to the y-axis is the integral over the solid of the function which gives the square of the distance of each point in the solid from the y-axis. It will be helpful to first label the corner points at the top of the region.
The resulting projection is shown in the figure above and to the right. The possible values of x depends on where we are in the yz-plane. Therefore, we split up the solid into two parts. The volume of the solid will be the sum of these two smaller volumes.
Figure 1: When the center of mass is in this position, it will go lower when a little more soda leaks out since mass above the center of mass is being removed. Figure 2: When the center of mass is in this position, it was lower moments before since mass that was below the center of mass was removed, causing the center of mass to rise. Therefore, the center of mass is lowest when it is at the height of the soda, as in Figure 3. The same argument would hold for a soda bottle. The region is a right circular cylinder about the z- The region is a hollow right circular cylinder 6.
The region is one-eighth of a sphere in the first about the z-axis with inner radius 1, outer radius octant of radius a, centered at the origin. See comment at beginning of write-up of Assume that the hemisphere lies above the xy- Problem 26 of the previous section. Assume that the hemisphere lies above the xy- Position the ball with its center at the origin.
Position the ball with its center at the origin and consider the diameter along the z-axis. Position the sphere above and tangent to the xy-plane at the origin and consider the point on the boundary to be the origin. See Problem 25b. Let m1 and m2 be the masses of the left and right balls, respectively. Jacobian 4. J u, v Problem Set Solving for x and y gives 9. The Jacobian is Solving for x and y The integral in the uv-plane is more 2 easily done by holding v fixed and integrating u.
Solving for x and y 7. Thus, 8. True: For each x, the density increases as y increases, so the top half of R is more b. The marginal PDF for U is obtained by dense than the bottom half. For each integrating over all possible v for a fixed y, the density decreases as the x u. True: See Section True: Use result of Problem 33, Section False: There are 6. False: The integrand should be r.
True: Inside integral is 0 since sin x3 y3 is an odd function in x. True: Use Problem 33, Section Each 2 2 integrand, e x and e2 y , determines Sample Test Problems and even function. Answers may vary. Since we are restricted to the Thus, 7. Now if Now if Note: this can be verified by finding the 1 distance between the points 0,9 and 9, 0. Now if 2t dt Related Papers. Instructor's Resource Manual.
By tanzila azis. Kalkulus Purcell's Solution. By milo nimo. Instructor's Resource Manual Section 0. By Adil Suprayitno. Calculo 9e Purcell-Varberg-Rigdon Solution. By Bryan Tircio. Section Download pdf. Remember me on this computer. Enter the email address you signed up with and we'll email you a reset link. Need an account? Click here to sign up.
Kalkulus dan Geometri Analitis, Jilid 1 (Edisi Kelima)
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