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Bladerunner1122
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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 yaxis as shown in the figure to the right.
 one year ago
 one year ago
Bladerunner1122 Group Title
Let R be the region in the first quadrant bounded by the graph y=3√x the horizontal line y=1, and the yaxis as shown in the figure to the right.
 one year ago
 one year ago

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Bladerunner1122 Group TitleBest ResponseYou've already chosen the best response.0
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 yaxis 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.
 one year ago

Bladerunner1122 Group TitleBest ResponseYou've already chosen the best response.0
I have no idea how to solve this one, so any and all help is appreciated.
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
dw:1362168263188:dw
 one year ago

Bladerunner1122 Group TitleBest ResponseYou've already chosen the best response.0
How is this supposed to work math wise? I'm having trouble visualizing how that becomes Calc.
 one year ago

Bladerunner1122 Group TitleBest ResponseYou've already chosen the best response.0
finally it's back up for me now.
 one year ago

Bladerunner1122 Group TitleBest ResponseYou've already chosen the best response.0
... so is that last part there the "actual" answer? I'm trying to wrap my head around your post.
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
does this make sense each plane is a sheet of volume x*x/2 * dy
 one year ago

Bladerunner1122 Group TitleBest ResponseYou've already chosen the best response.0
Yeah, I think so. Thanks.
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
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= (3y)^2 \[ \frac{1}{2} \int_{1}^{3} (3y)^4 \ dy\]
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago
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