# Let R be the shaded region in the first quadrant enclosed by the y-axis and the graphs of y=4-x^2 and y=1+2sinx, how do you find the volume of the solid whose base is R and whose cross sections perpendicular to the x-axis are squares?

Jul 12, 2016

$= {\int}_{x = 0}^{1.102} \setminus {\left(3 - {x}^{2} - 2 \sin x\right)}^{2} \setminus \mathrm{dx}$

#### Explanation:

at any given x, the cross section area of the square will be

$A \left(x\right) = {\left({y}_{2} - {y}_{1}\right)}^{2}$ {where ${y}_{2}$ is the green line in the attached plot}

with ${y}_{2} = 4 - {x}^{2}$

and ${y}_{1} = 1 + 2 \sin x$

so volume V(x) follows as

$V \left(x\right) = {\int}_{x = 0}^{1.102} \setminus A \left(x\right) \setminus \mathrm{dx}$

$= {\int}_{x = 0}^{1.102} \setminus {\left(3 - {x}^{2} - 2 \sin x\right)}^{2} \setminus \mathrm{dx}$

Please note that I got the 1.102 off the Desmos plot included here. I did not solve it myself.

Also that integral is horrendous so I would recommend a computer solution if you have access. This is really not something you want to be doing by hand.