# Thread: Finding area of a region

1. ## Finding area of a region

Would that give me the area for S? The formula that I took a picture of.
Im doing (b)

2. ## Re: Finding area of a region

I dont think so.

This gives you the right soln

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

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)

3. ## Re: Finding area of a region

I think so..............
u would be right ..............

subtrat from

to get

5. ## Re: 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.

6. ## Re: Finding area of a region

Guys, he is asking about The area of the region S, not the volume obtained from revolution.

7. ## Re: 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 $S=\int_3^4 (18-3x)-(4x^2-x^3)\,dx + \int_4^6 18-3x\,dx$.

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 $\int_0^4 \pi(4x^2-x^3)^2\,dx$.

- Hollywood