posted by Shay
Hello. I would appreciate it if someone could check my answers. I'm sorry it is so long.
1.) Let R denote the region between the curves y=x^-1 and y=x^-2 over the interval 1<= x <= 10.
a. Set up an integral for the area of R.
My answer: 1.403
b. Find x-bar, the x coordinate of the centroid of R.
My answer: 4.775
c. Set up and evaluate an integral for the volume of revolution of the solid generated when R is revolved about
i. the x-axis
My answer: 1.781
ii. the y-axis
My answer: infinity
2.) The length of a cable is 50 and the weight is 10. A portion of length 40 was hanging over the edge of a tall building and was pulled to the top. How much work was done?
My answer: 3920
3.) Let C denote the curve y= x(4-x), where 0<= x <= 4. Set up the integral for the following. In this case, do not evaluate the integrals.
a. the length of C
My answer: integral from 0 to 4 of sqrt[ 1+ (4-2x)^2] dx
b. the area of the surface generated when C is revolved about
i. the x-axis
My answer: 2pi *integral from 0 to 4 of x(4-x)*sqrt[1+ (4-2x)^2] dx
ii. the y-axis
My answer: 2pi* integral from 0 to 4 of [sqrt(4-y) -2] *sqrt[1+ (-2*sqrt(4-y))^-2] dy
4.) A tank has the shape of a trapezoidal prism. The top is horizontal and the two ends are vertical. The length is 4. The height is 2. The top is a 3-by-4 rectangle. Viewed from an end, the tank looks like the trapezoid shown in the figure below. Assume the tank contains a liquid to a depth of 1. Take the density of the liquid to be p.
a. Set up, but do not evaluate, an integral for the work required to pump the liquid to the top of the tank.
My answer: W= p*integral from 0 to 2 of 4 dy
b. Set up, but for not evaluate, an integral for the fluid force against one end of the tank.
My answer: F= integral from 0 to 4 of p(24) dx
One note: to check your integrals,
1a. Correct b. agree c.
c. agree d. Rotating about y...it has a beginning radius and an ending radius in the x domain, so the area of revolution is bounded. It has a bounded volume of revolution. Rethink your limits. You probably will have to break up the integral to three volumes, a bottom, middle cylinder, and a top portion.
2. The integral must be weight * dheight. weight for length dx is 10/50 ...no 9.8 is needed. So work= height *d weight* = INT y 10/50dy= 10/50 y^2/2 limits 0,40
Now think to check this: the weight is .4*10, the average height is 40/2, which checks with the above.
3. a. correct i. yes ii. you have dy limits 0,4...I don't understand that. Secondly, rotating about the y axis, the outer radius is 4, the inner radius is y.
4. I don't understand this. Make a thinkness dy, figuure the area as a function of y. Then the volume is area*dy, and weight is rho*area*dy. Let y=0 be the top of the water.
then you raise the water 1+y
work= weight*height. You can check by figuring the total weight, and the average height is raised.
For number 1d I don't understand what the limits would be.
For 1d. Draw a picture of the two functions, make your area, and rotate about y. If you wish, you can post your drawing on the web again.
What would the integral for number 4 be?
Find the slope and y-intercept of each line...