Would that give me the area for S? The formula that I took a picture of.

Im doing (b)

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- Feb 7th 2013, 07:56 PMasilvester635Finding area of a region
Would that give me the area for S? The formula that I took a picture of.

Im doing (b) - Feb 7th 2013, 08:28 PMjakncokeRe: Finding area of a region
I dont think so.

This gives you the right soln

$\displaystyle \int_3^6 y dx - \int_3^4 f(x) dx $

Basically get the area under the curve of the line from where the line is tangent to f(x)(x=3) to where the line intersects the x-axis and subtract that from, the point where the line is tangent to f(x) (x=3) to where f(x) intersects the x-axis (x=4) - Feb 7th 2013, 08:39 PMpssingh1001Re: Finding area of a region
I think so..............

u would be right .............. - Feb 7th 2013, 08:43 PMjakncokeRe: Finding area of a region
- Feb 7th 2013, 08:45 PMibduttRe: Finding area of a region
You are absolutely right. Area R would be integral from 0 to 4 of f(x).

Volume of revolution would be 2pi integral of square of f(x) from 0 to 4. - Feb 7th 2013, 08:56 PMjakncokeRe: Finding area of a region
Guys, he is asking about The area of the region S, not the volume obtained from revolution.

- Feb 7th 2013, 09:30 PMhollywoodRe: Finding area of a region
The tangent point is x=3, f(x) crosses the x-axis at x=4, and the line crosses the x-axis at x=6. So if you divide the area S into two regions with a vertical line at x=4, you have $\displaystyle S=\int_3^4 (18-3x)-(4x^2-x^3)\,dx + \int_4^6 18-3x\,dx$.

And I think that gives the same answer as jakncoke.

I think there's an extra 2 in ibdutt's answer to part (c) - it should be $\displaystyle \int_0^4 \pi(4x^2-x^3)^2\,dx$.

- Hollywood