Sample Midterm I Solutions

# Sample Midterm I Solutions - Solutions to the 1st Midterm 1...

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Solutions to the 1st Midterm 1. Consider the following integral Z 1 0 Z 1 - 2 x - 1 Z 1 - y 2 0 f ( x, y, z ) dz dy dx. (a) Rewrite this integral so that the inner-most integral is with respect to z , the middle integral is with respect to x , and the outer integral is with respect to y . As given in the integral, the region of integration can be viewed as the region under the graph of z = 1 - y 2 over a domain D in the xy -plane. Further, D is the domain bounded by the lines y = - 1, x = 0, and y = 1 - 2 x . If we want to switch the order of x and y integration, this means that y goes from - 1 to 1 while x goes from 0 to 1 2 (1 - y ). Thus the integral can be re-written as Z 1 - 1 Z 1 2 (1 - y ) 0 Z 1 - y 2 0 f ( x, y, z ) dz dx dy. (b) Rewrite this integral so that the inner-most integral is with respect to x , the middle integral is with respect to z , and the outer integral is with respect to y . In this case, we think of our region as the region under a graph of x as a function of y and z over some domain in the yz -plane. The domain in the yz -plane is bounded by z = 0 and z = 1 - y 2 , so y goes from - 1 to 1 while z goes from 0 to 1 - y 2 . In the x -direction, the region is bounded form below x = 0 and from above by x = 1 2 (1 - y ). Thus the integral can be re-written as Z 1 - 1 Z 1 - y 2 0 Z 1 2 (1 - y ) 0 f ( x, y, z ) dz dx dy. 2. Consider the surface given by the part of the plane 2 x + y + 2 z = 8 that lies in the ﬁrst octant. Find its surface area. Be sure to show the integral you set up as well as your ﬁnal answer. The part of the plane which lies in the ﬁrst octant is a triangle with vertices at (4 , 0 , 0), (0 , 8 , 0), and (0 , 0 , 4); these are just the intersections of the plane with the coordinate axes. We can view this as the graph of z = 4 - x - 1 2 y over the triangle in the xy -plane bounded by x = 0, y = 0 and 2 x + y = 8. To ﬁnd the area of the graph of a function, we integrate p ( ∂f/∂x ) 2 + ( ∂f/∂y ) 2 + 1 over the domain. In the present case, we have that

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## This test prep was uploaded on 04/07/2008 for the course MATH V1202 taught by Professor Neel during the Fall '07 term at Columbia.

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Sample Midterm I Solutions - Solutions to the 1st Midterm 1...

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