Calculus 2 volume integration, y=axis

funkyguy4000
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Hello, so I've been having lots of trouble with a math problem.
I can differentiate functions but I'm kinda confused on what to do when it is like y=4

So say you have the given: y= x^2 and then y=4, x=0, x=2. Rotate about the y-axis.
I'm know you have to get it into respect to y so x=sqrt(y) but what would I do from there?! you can't put x in terms of y for y=4! I'm confused. Can somebody help me out or point me in the right direction.
 
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they are giving you 4 different curves or lines and you are supposed to envision the region bounded by those curves or lines. so you have the line y=4 at the top, y=x^2 on the right, ...

ooops, your equations do not fit together to give one region. you need to re - read the question.
 
Okay well the y=x^2 is a parabola so it goes on both sides of the y-axis. And then the y=4 is another bounding line.

we are supposed to rotate the bounded region from x=0 to 2 around the y-axis.

I'm just confused on how to rotate it around like that.
 
well you don't want to say x=2 if you know y=4 and y = x^2. rather you should say y = x^2 for x≥0, y=4, and x=0. but if that's what it said, its their fault not yours.

so apparently the region is bounded by y = x^2 for x≥0, x = 0 and y=4. that makes it a portion of the right half of the parabola. then you rotate it around the y axis. is that hard to see? your book may not be too well written.
 
well they give me the limits x=0 and x=2. y=x^2 and y=4 intersect at x=2.

I can imagine it. I'm just confused how to rotate it around the y=axis. what method to use and how to get the variables needed in the method. I tried the shells method and that didn't work. I didn't understand how to get the needed material.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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