Reverse Integrals. (Check my work?)

**Tall Jessica** · November 25th 2010, 11:19 PM

I'm doing some work - yes, on Thanksgiving

- and I've been doing good, but the paper reversed things on me! lol... I have to BUILD an integral, not solve it! Anywho, please check my work!

The problem(s) state(s):
Set up, but do not evaluate, for the volume of the solid obtained by rotating the region bounded by the given curves around the specified axis.

1) $y=sinx$ , $y=0$ , $x={2}\pi$ , $x={3}\pi$ ; about the y-axis

2) $x=\sqrt{sin y}$ , $0 \leq y \leq \pi$ , $x=0$ ; about $y=5$

Okay, so for the first problem:

Looking at the graph and rotating it, I get a nice little donut. Since I need my approximating squares to touch BOTH functions, I have to use the Shell Method......right?

$\int_{{2}_\pi}^{{3}^\pi}2\pi{r}{h})dx$

So, I would take my $h = sin x$ and my $r = x$ , giving me:

$\int_{{2}_\pi}^{{3}^\pi}2\pi({sin x})({x})dx$

Right?

For the second problem, it's the same formula, but with $r=5-y$ and $h=\sqrt(siny)$ , giving me:

$\int_{0}^{\pi}2\pi({5-y})({\sqrt(siny)})dx$

Thoughts?

**HallsofIvy** · November 26th 2010, 04:00 AM

Originally Posted by Tall Jessica

I'm doing some work - yes, on Thanksgiving

- and I've been doing good, but the paper reversed things on me! lol... I have to BUILD an integral, not solve it! Anywho, please check my work!

The problem(s) state(s):
Set up, but do not evaluate, for the volume of the solid obtained by rotating the region bounded by the given curves around the specified axis.

1) $y=sinx$ , $y=0$ , $x={2}\pi$ , $x={3}\pi$ ; about the y-axis

2) $x=\sqrt{sin y}$ , $0 \leq y \leq \pi$ , $x=0$ ; about $y=5$

Okay, so for the first problem:

Looking at the graph and rotating it, I get a nice little donut. Since I need my approximating squares to touch BOTH functions, I have to use the Shell Method......right?

$\int_{{2}_\pi}^{{3}^\pi}2\pi{r}{h})dx$

So, I would take my $h = sin x$ and my $r = x$ , giving me:

$\int_{{2}_\pi}^{{3}^\pi}2\pi({sin x})({x})dx$

Right?

yes, that is correct- well, except that your upper limit on the integral should be $3\pi$ , not $3^\pi$ !

For the second problem, it's the same formula, but with $r=5-y$ and $h=\sqrt(siny)$ , giving me:

$\int_{0}^{\pi}2\pi({5-y})({\sqrt(siny)})dx$

Yes, that is correct.

Thoughts?

**Tall Jessica** · November 26th 2010, 04:17 AM

Thank you! I am aware that the upper limit is $3\pi$ , but I am new to LaTeX! Thank you so much!

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