• Support PF! Buy your school textbooks, materials and every day products Here!

Mass of Region Bounded by y=sin(x), z=1-y, z=0, and x=0

  • #1

Homework Statement


On a sample midterm for my Calc 3 class the following question appears:

Find the mass of (and sketch) the region E with density ##\rho = ky## bounded by the 'cylinder' ##y =\sin x## and the planes ##z=1-y, z=0, x=0## for ##0\le x\le\pi/2##.

Homework Equations


$$ m= \int_{E} \rho dV$$

The Attempt at a Solution


By projecting down to the x-y plane the region is given by ##(x,y,z): 0\le x\le\pi/2, \sin x\le y\le1, 0\le z\le1-y##.

So the mass of the region is $$\int_{0}^{\pi/2} \int_{\sin x}^1 \int_{0}^{1-y} ky dzdydx$$.

Performing all the iterated integrations gives ##k(\frac{2}{9} -\frac{\pi}{24})## but the stated answer is ##k(\frac{16-\pi}{72})##.
 
Last edited by a moderator:

Answers and Replies

  • #2
33,644
5,311

Homework Statement


On a sample midterm for my Calc 3 class the following question appears:

Find the mass of (and sketch) the region E with density ##\rho = ky## bounded by the 'cylinder' ##y =\sin x## and the planes ##z=1-y, z=0, x=0## for ##0\le x\le\pi/2##.

Homework Equations


$$ m= \int_{E} \rho dV$$

The Attempt at a Solution


By projecting down to the x-y plane the region is given by ##(x,y,z): 0\le x\le\pi/2, \sin x\le y\le1, 0\le z\le1-y##.

So the mass of the region is $$\int_{0}^{\pi/2} \int_{\sin x}^1 \int_{0}^{1-y} ky dzdydx$$.

Performing all the iterated integrations gives ##k(\frac{2}{9} -\frac{\pi}{24})## but the stated answer is ##k(\frac{16-\pi}{72})##.
The two expressions are equal. Combine the two terms in the first expression using a common denominator of 72.
 
  • #3
Oh, well then my terrible algebra skills strike again. Thanks
 

Related Threads on Mass of Region Bounded by y=sin(x), z=1-y, z=0, and x=0

Replies
3
Views
663
Replies
2
Views
1K
  • Last Post
Replies
2
Views
5K
Replies
7
Views
8K
Replies
1
Views
1K
Replies
6
Views
1K
Replies
3
Views
9K
Replies
3
Views
1K
Top