Part I:
Hand in the underlined problems; the others are for practice.
Lecture 16. Thurs. (March 11): Double and iterated
integrals in rectangular coordinates.
Read EP 14.1-14.3. Concentrate on:
Lecture 18. Tues. (March 16): Double integrals in polar coordinates; applications.
Problem 1. (Thurs. 3 pt) A rectangular prism is made by taking a long piece of wood with a rectangular cross-section, sawing off one end perpendicularly, and the other end at an arbitrary angle (so that the four edges have in general four different lengths). Show by double integration that
volume of prism (cross-sectional area) (average of the four lengths).
[Place the prism as shown in the picture, and take as the equation of the top surface.] Oops, fogot the picture. Oh well I hope is was reasonably clear anyway.
This formula can be thought of as generalizing the formula for the area of a trapezoid.
Solution: If the prism is placed as suggested, with sides in the three coordinate planes and sloping end of the form then the volume of the prism is the integral over the rectangular region of the xy-plane where of
Problem 2. (Tues. 3 pt) Work Problem 32 EP page 901.
Solution: Change the order of integration - the region is a triangle and
Problem 3. (Tues. 3 pt) A split log has a semi-circular cross-section, and a radius . A wedge-shaped piece is cut out of it. What is the volume of the piece? ( Hint: Calculate half the volume of something else.)
Solution: There is likely to be much confusion about this one, because I did not add a picture. Give full marks for any reasonable interpretation. I meant that the top surface of the log was and it lies in stretching in the y-direction. The prism was supposed to be in the xy-plane, in - so between the lines and The volume is therefore twice the integral over of i.e.
As I say, interpretation may differ on this one!
Problem 4. (Tues. 3 pt) Work Problem 18, EP page 913.
Solution: Change to polar coordinates, the main problem is to find the limits of integration. The region is restricted by The condition means This is the circular region with center and radius Thus the region of integration is a quarter of this disk - the upper right quarter. So the whole integral becomes, in polar coordinates
Problem 5. (Tues. 3 pt) Find the centroid (center of gravity) of the plane lamina lying between the parabolas and , if the density function is .
Solution: By symmetry the centroid must lie on the x-axis, so we only need to compute - we can compute this for the upper half, since it will have the same We can set since it will cancel, thus
Mass |
Problem 6. (Tues. 3 pt) Find the moment of inertia of a thin circular plate of radius and density about
(Hint: In both cases place the point at the origin and place the circle so a diameter lies along the -axis. Use polar coordinates. There are integral tables in the front and back cover of your book which use can use.)
Solution: