Math 30G: Calculus Homework 12
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1 Math G: Calculus Homework ue b Frida, Oct. 9, (6 pm O NOT E-ISTIBUTE THIS SOLUTION FILE. Let be the region inside the unit circle centered at the origin, let be the right half of, and let B be the bottom half of. ecide, without calculating the value of an of the integrals, whether each integral is positive, negative, or zero. (a da (b (c (d (e B 5 da 5 da e da e da B (a The double integral da is positive, since it is the area of the region (which is the area of a unit circle, namel π. (b The value of 5 da is positive since the integrand 5 is greater than or equal to over all the points in the region (i.e., -coordinate is greater than or equal to for all points (,. (c The double integral 5 da is zero since region B is smmetric across the -ais (i.e., B smmetric with respect to and 5 is an odd function, meaning contributions to the double integral from the left of -ais will cancel with contributions from the right. (d The value of e da is positive since the integrand, e, is positive over all points (, in the region. (e The double integral e da positive. Notice that the integrand e is an increasing function for. Therefore, while the area of the half of region on the left of -ais is the same as the area of the half on the right of -ais and the integrand is negative on the Math G - Prof. Kindred
2 left and positive on the right, the values of the integrand are larger in magnitude (absolute value on the right than the are on the left.. Consider the integral ( + d d. (a Sketch the region over which the integration takes place. (b Write an equivalent integral with the order of integration reversed. (Note: ou do not need to compute the integral. (a The region of integration is shown as the shaded blue region below: (b If we integrate with respect to first, then the boundaries of the region with respect to are to. Furthermore, we follow with the boundaries for as to. Thus, switching the order of integration to d d ields the following double integral: ( + d d.. Find sin where is the triangle in the -plane bounded b the -ais, the line and the line. (Note: ou ma want to think a bit about which order of integration will be easier to compute. Just looking at the integrand, we realize that we do not have a wa of finding the antiderivative of sin if we first integrate with respect to. So we would like to integrate with respect to first. To find our limits of integration, we sketch the region of integration,. da Math G - Prof. Kindred
3 Thus, we have sin d d sin sin d d ( cos cos + cos cos.. Evaluate da where is the region bounded b the line and the parabola The region for Question. Based on the region of integration, we choose to integrate first with respect to. Thus, the boundaries of the region in the -direction are given b 6 and +. Then to find the range of values for, we need to find the two -coordinates in the points of intersection between the curves and 6. We substitute + into the second equation to obtain ( ( ( +,. Math G - Prof. Kindred
4 Therefore, the double integral is da + 6 d d [ ( d ( ( + ] 6 d ( d ( d ( d + + [( ( ( 6 [( 8 6 ( ] + ( + ( ( ( 6 8 ] Evaluate the integral 9 sin( d d Based on the integrand, sin(, we want to switch the order of integration to d d. Note that in the -direction, the boundaries of the region of integration (shown above as blue shaded region are to. We see then that ranges from to 9. Thus, switching the order of integration Math G - Prof. Kindred
5 ields 9 sin( d d 9 9 sin( sin( d. d We can make use of u-substitution, letting u, so that du d. The limits of integration are now u and u 9 8, so we have 9 sin( d 8 sin(u du [ cos(u] 8 cos 8 (cos cos 8. Note: the solution is also equal to ( 8 sin. 6. Evaluate W the first octant. dv where W is the region under the plane + + z 6 that lies in We first sketch the region W : 6 plane z z If we integrate first with respect to z, then the boundaries of the region in the z-direction are z and z 6. Having handled the z-direction, we project W down into the - plane to obtain the two-dimensional region at right. (Note that the equation of the line is found b letting z in the equation for the plane. We choose net to integrate with respect to, and we see that the boundaries are now given b + 6 and 6. Finall, we see that ranges from to. Now that we have found the limits of integration, we can 5 Math G - Prof. Kindred
6 set up the triple integral and evaluate it: W dv [ z6 z z z6 z dz d d d d (6 d d ( 6 d d 6 [( ] d [ ( ] ( (6 6 d ] [ d [ ] 8 + d ] ( Math G - Prof. Kindred
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