# Thread: Help setting up integral involving shell/washer method

1. ## Help setting up integral involving shell/washer method

I need help with the following problem:

Consider the region T bounded by x=(y^3)+3, x=45-y,y=1, and y=3.

I am allowed to use the shell or washer method, whichever works.

a.) I need help setting up an integral required to find the volume of the solid obtained by revolving the region T about the line y=-1.

b.) I need help setting up an integral required to find the volume of the solid obtained by revolving the region T about the line x=50.

I'm not really sure how to even begin setting up a. I think to set up b you would use:

45
2PI∫ (50-x)(y^3 +3 -3) using the shell method. Have I set this
5

integral up right or have I done something wrong?

I only need help figuring out how to set up the integrals. Thanks to anyone who can help.

2. a. shells ... radius of rotation is $r = y - (-1) = y+1$

$V = 2\pi \int_1^3 (y+1)[(45-y)-(y^3+3)] \, dy$

b. washers ... $R = 50-(y^3+3) = 47-y^3$
$r = 50 - (45-y) = 5+y$

$V = \pi \int_1^3 (47-y^3)^2 - (5+y)^2 \, dy$

3. Hey guys. Been working on a routine to illustrate better, volumes of revolutions. This is what I think the first one looks like to me. The cone shape is the line $y=\sqrt[3]{x-3}$ from $x=4$ to $x=30$. Note, the equation $x=y^3+3$ is clipped by the boundary line $y=3$ at $x=30$. The red line is the axis of revolution. The inner radius is the other boundary line $y=1$. Anyway, I think a good way to improve your skills in solving these is to visualize the volume precisely. Also, the very act of writing the code to generate these is also helpful towards visualizing them. Once you got it accurately visualized, then the calculus follows easily.