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

In summary, the question is asking to find the mass of a region E with density ky, bounded by a cylinder and three planes. The answer is found by performing iterated integrations and simplifying to get the expression k(16-π)/72. However, this is equal to k(2/9-π/24) and can be written in a common denominator of 72.
  • #1
TeslaCoil137
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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})##.
 
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  • #2
TeslaCoil137 said:

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
 

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

What is the equation for the region bounded by y=sin(x), z=1-y, z=0, and x=0?

The equation for the region bounded by y=sin(x), z=1-y, z=0, and x=0 is z = 1 - sin(x). This can be obtained by setting the two equations, z=1-y and y=sin(x), equal to each other and solving for z.

What is the volume of the region bounded by y=sin(x), z=1-y, z=0, and x=0?

The volume of the region bounded by y=sin(x), z=1-y, z=0, and x=0 is . This can be calculated by setting up a double integral and integrating with respect to x and y, with the limits of integration being 0 and 1 for both variables.

What is the surface area of the region bounded by y=sin(x), z=1-y, z=0, and x=0?

The surface area of the region bounded by y=sin(x), z=1-y, z=0, and x=0 is . This can be calculated by setting up a surface integral and integrating with respect to x and y, with the limits of integration being 0 and 1 for both variables.

How does changing the value of z=1-y affect the shape of the region bounded by y=sin(x), z=1-y, z=0, and x=0?

Changing the value of z=1-y will affect the shape of the region bounded by y=sin(x), z=1-y, z=0, and x=0 by shifting the region up or down along the z-axis. This will change the height of the region and therefore the volume and surface area as well.

What would be the result if the bounds of the region were changed to y=cos(x), z=1-y, and x=0?

The result would be a different region bounded by y=cos(x), z=1-y, and x=0. This region would have a different shape and therefore different volume and surface area. The equation for this region would be z = 1 - cos(x).

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