# Shell Method

#### alane1994

##### Active member
Here is my homework question. I am stuck on one part of it, and it is ok for me to receive guidance, not answers without effort.

Question:
Let R be the region bounded by the following curves. Use the shell method to find the volume of the solid generated when R is revolved about the x-axis.

$$y=\sqrt{98-2x^2}$$

My Work So Far:
• I have found the value of x
$$x=\pm\sqrt{\frac{98-y^2}{2}}$$
• The lower limit is 0.
• The upper limit is___?

This is where I get stuck... I am unsure how to get the upper limit. Once I get that, I should be able to proceed from there.

#### tkhunny

##### Well-known member
MHB Math Helper
#1 - Please observe symmetry. Work in the 1st Quadrant and multiply by 2 to achieve the entire result. This will free you from the laborious "+/-".

Thus: $$2\cdot\left(2\cdot\pi\cdot\int_{0}^{\sqrt{98}}y \cdot x\;dy\right)$$

You've only to substitute your correct expression for 'x' and you're done.

In my opinion, you should ALWAYS do it the other way in addition to what is asked. This will do at least these three things:
1) Give you experience in both methods.
2) You will gain experience in judging which is better in which circumstances.
3) You will be able to check your own work.

$$2\cdot\left(\pi\cdot\int_{0}^{7}y^{2}\;dx\right) = 2\cdot\left(\pi\cdot\int_{0}^{7}98 - 2x^{2}\;dx\right)$$

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