ozgunozgur
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- Apr 1, 2020
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You need to submit an attempt at solving the problem, and we can help you where you are stuck. In the mean time, I can get you started.
In both parts, start by making a graph of the 3 equations and shade the region that is bounded by them.
Then both questions can be solved using the disk method for volume.
You found the anti-defivative of the integrand, but you need to evaluate a definite integral.
No, the strips run from \(x=0\) to \(x=2\). We are given the lower bound, since the vertical line \(x=0\) is one of the boundaries of the shaded region. The upper limit fomrs from there the quadratic and linear functions intersect:
[MATH]x^2+4=4x[/MATH]
[MATH]x^2-4x+4=0[/MATH]
[MATH](x-2)^2=0\implies x=2[/MATH]
Hence:
[MATH]V=\pi\int_0^2x^4-8x^2+16\,dx=\frac{256}{15}\pi[/MATH]
When revolving about the \(y\)-axis, I would use the shell method, where:
[MATH]dV=2\pi x((x^2+4)-4x)\,dx[/MATH]
Do you see that the radius of an arbitrary shell is \(x\) and the height is the difference between the quadratic and the line?
Please Google those terms:I found 8 pi. Is this correct? What is the difference between washer and shell method?
I found 8 pi. Is this correct? What is the difference between washer and shell method?