Let R be the region in the first quadrant bounded by the graph y=3-√x the horizontal line y=1, and the y-axis as shown in the figure to the right.

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Let R be the region in the first quadrant bounded by the graph y=3-√x the horizontal line y=1, and the y-axis as shown in the figure to the right.

Calculus1
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http://goo.gl/AhgJX Region R is the base of a solid. For each y, where 1≤y≤3, the cross section of the solid taken perpendicular to the y-axis is a rectangle whose height is half the length of its base. Write, but do not evaluate, an integral expression that gives the volume of the solid.
I have no idea how to solve this one, so any and all help is appreciated.
  • phi
|dw:1362168263188:dw|

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How is this supposed to work math wise? I'm having trouble visualizing how that becomes Calc.
finally it's back up for me now.
... so is that last part there the "actual" answer? I'm trying to wrap my head around your post.
  • phi
does this make sense each plane is a sheet of volume x*x/2 * dy
Yeah, I think so. Thanks.
  • phi
each plane is a sheet of volume x*x/2 * dy now put everything in terms of all x or all y if we stick with x: y=3-√x and \[dy = -\frac{1}{2} x^{-\frac{1}{2}} \ dx\] the limits of integration will be x=4 to 0 \[ -\frac{1}{4} \int_{4}^{0} x^{\frac{3}{2}}\ dx\] or flipping the limits, and negating \[ \frac{1}{4} \int_{0}^{4} x^{\frac{3}{2}}\ dx\] if we go with y: x= (3-y)^2 \[ \frac{1}{2} \int_{1}^{3} (3-y)^4 \ dy\]
  • phi
fixed a typo on finding how dy equates to dx. the first way changes dy to a function of x and dx and then integrates over x the second way changes x to a function of y and integrates the volume x^2/2 (in terms of y) over y

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