Help with finding the area of a double integral bounded by a region

ktalltheway

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1698032742849.png For this question I keep getting the answer of 0 even though I know that isn't right. The two parabolas perfectly mirror one another so should i just set the lower limit of the y integral to y=0 then multiply the final derivative I get by 2? I know that's not how I'm supposed to do it, but it feels a nice shortcut to this problem that I have been stuck on forever.
 
View attachment 36609 For this question I keep getting the answer of 0 even though I know that isn't right. The two parabolas perfectly mirror one another so should i just set the lower limit of the y integral to y=0 then multiply the final derivative I get by 2? I know that's not how I'm supposed to do it, but it feels a nice shortcut to this problem that I have been stuck on forever.

Please reply *showing* your work. Thank you!
 
View attachment 36609 For this question I keep getting the answer of 0 even though I know that isn't right. The two parabolas perfectly mirror one another so should i just set the lower limit of the y integral to y=0 then multiply the final derivative I get by 2? I know that's not how I'm supposed to do it, but it feels a nice shortcut to this problem that I have been stuck on forever.
No! The bounded region may have symmetry but does the height, 6xy5, have this symmetry as well??? Remember you are finding the volume above the given region. Where does the height come from?
Even if you are an expect in solving these problems in my opinion it doesn't mean much if you don't understand what the integral is asking. So please think about what is being asked. For example, if the answer is 18, then what does that answer even mean?
 
View attachment 36609 For this question I keep getting the answer of 0 even though I know that isn't right. The two parabolas perfectly mirror one another so should i just set the lower limit of the y integral to y=0 then multiply the final derivative I get by 2? I know that's not how I'm supposed to do it, but it feels a nice shortcut to this problem that I have been stuck on forever.
Can you integrate the given function over one region - say y = x2 - 1? Please share your work.
 
View attachment 36609 For this question I keep getting the answer of 0 even though I know that isn't right. The two parabolas perfectly mirror one another so should i just set the lower limit of the y integral to y=0 then multiply the final derivative I get by 2? I know that's not how I'm supposed to do it, but it feels a nice shortcut to this problem that I have been stuck on forever.
How do you know 0 isn't right? Who told you that?

And why haven't you shown the work that gave you that answer? Maybe it's actually right; or maybe we could point out a mistake.

There is symmetry in both the region (more than you mention) and in the integrand. Those will interact in a way different from what you suggest. And you may well be expected to use this as a shortcut. You just have to do it right.
 
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