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

## 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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## 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.

Oh, well then my terrible algebra skills strike again. Thanks