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Use the reduction formula for the integral: $$\displaystyle I_{a,n}= \int_{0}^{1} u^ne^{au} du, a\in\mathbb{R} ,a\neq0,n=0,1,2,\dots $$ to calculate the volume of the solid of revolution obtained by revolving the region between the graph of the function $x=e^{\sqrt[3]{y}}$ and the y-axis, above the interval $[0,1]$ about

$a)$ The line $x=-1$.

$b)$ The line $y=-1$.

$\textbf{MY ATTEMPT}$:

The reduction formula was given to me. It's just:

$I_{a,0}=\frac{1}{a}(e^{a}-1)$

I'm confused about the "above the interval $[0,1]$". Is that $x$ or $y$? How would you know?

If it's $x$ I can't see how this is possible since the graph has an vertical asymptote at $x=0$ meaning that the area "under" it is infinite.

Second, I'm not if I should use the washer method or the shell method to do this. I did plot this and it LOOKS like this is a shell, but I am not sure...

enter image description here

For part a, it definitely looks like a shell and for part b, it looks like a washer, but I am not sure.

Can someone help me set up the integrals?

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1 Answer 1

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For the first one, that is volume around $x=-1$ axis, using disk method, we have

$$V_1=\pi\int_0^1 \left(e^{\sqrt[3]{y}}-(-1)\right)^2\;dy-\pi\int_0^1 \left(0-(-1)\right)^2\; dy$$

enter image description here

for the second one around $y=-1$ axis, using shell method, we have

$$V_2=2\pi\int_0^1 (y-(-1))e^{\sqrt[3]{y}}\;dy$$

enter image description here

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  • $\begingroup$ For $V_1$, did you mean to square the 2nd integral? Not that the answer will change but... $\endgroup$
    – Future Math person
    Commented Jan 21 at 10:57
  • $\begingroup$ Yes of course! I fix that $\endgroup$
    – user
    Commented Jan 21 at 11:03
  • $\begingroup$ Feel free to ask for any clarification $\endgroup$
    – user
    Commented Jan 21 at 11:05

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