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I have to calculate an integral, from which I am using the method Slicing, and Cylindrical Shells. I was thus rotating it by the x-axis.

I am using the functions $x=\frac{y}{4}$, and $x=\sqrt{y}$

I did it using slicing, and get this integral, and the answer.

$$V_1=\pi\int_\limits{0}^{4}((4x)^2-(x^2)^2)dx$$ This is then later equal to $V_1=\frac{2048}{15}\pi$ Then using cylindrical Shells method to get the answer:

$$V_2=2\pi\int_\limits{0}^{16}(y(\frac{y}{4}-\sqrt{y}))dy$$ And it does not equal the same $V_1$. How does one do it using cylindrical shells?

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The way you have set up the integral for the washer method is correct. However, the integral for the shell method (or the cylindrical shell method as it's also called) is slightly wrong. It should be the following:

$$ V_2=2\pi\int_{0}^{16}\bigg[y\left(\sqrt{y}-\frac{y}{4}\right)\bigg]\,dy= 2\pi\bigg[\frac{2}{5}y^{5/2}-\frac{1}{12}y^3\bigg]_{0}^{16}= \frac{2048\pi}{15} $$

You should be subtracting $\frac{y}{4}$ from $\sqrt{y}$ because $\sqrt{y}$ lies to the right of $\frac{y}{4}$ or above it if you rotate the entire coordinate system 90° to the right and then flip it so that it looks like x and y have exchanged placed.

Wolfram Alpha gives the same answer.

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If you evaluate your second integral accurately, the result (multiplying by $2\pi$) comes to $-\frac{2048}{15}\pi$. The reason it is different from the other result is you subtracted the larger value of $y$ from the smaller value of $y.$ You should subtract the smaller value from the larger one.

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over the interval $[0,16], \frac {y}{4} < \sqrt y$

You have the sign backwards.

$2\pi \int_0^{16} y(y^{\frac 12}- \frac{y}{4})\ dy\\ 2\pi \int_0^{16} y^{\frac 32}- \frac{y^2}{4})\ dy\\ 2\pi(\frac {2}{5} y^{\frac 52}- \frac {y^3}{12})|_0^{16}\\ 2\pi(4^5)(\frac 1{15})$

And washers:

$\pi\int_0^4 16 x^2 - x^4\ dx\\ \pi(\frac {16x^3}{3} - \frac {x^4}{5})|_0^4\\ \pi(4^5)(\frac 2{15})$

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