Double Integral, how to find volume of region ?

[mrjust]

Third semester calculus; What I'm I doing wrong? This is for my online homework and it's marking my answer wrong.

My problem:

Find the volume of the region under the graph of f(x,y)= 5x+y+1 and above the region y^2<=x, 0<=x<=16.

My attempt:

First I drew the diagram for my bounds :



From here I setup the iterated integral (double integral) with the bounds I found ( wolframalpha link ):

http://www.wolframalpha.com/input/?i=int (5x+y+1)dydx, x=0 to 16, y=0 to sqrt(x)&t=ff3tb01

First I integrate with respect to y by holding x constant, then I integrate with respect to x and get the answer of 6464/3 cubic units.

 

[HallsofIvy]

Third semester calculus; What I'm I doing wrong? This is for my online homework and it's marking my answer wrong.

My problem:

Find the volume of the region under the graph of f(x,y)= 5x+y+1 and above the region y^2<=x, 0<=x<=16.

This makes no sense. This region is two dimensional, not three dimensional, so has area, not volume. Are you rotating this region around an axis?

My attempt:

First I drew the diagram for my bounds :

View attachment 3978

From here I setup the iterated integral (double integral) with the bounds I found ( wolframalpha link ):

http://www.wolframalpha.com/input/?i=int (5x+y+1)dydx, x=0 to 16, y=0 to sqrt(x)&t=ff3tb01

First I integrate with respect to y by holding x constant, then I integrate with respect to x and get the answer of 6464/3 cubic units.

 

[mrjust]

I figured it out I had the incorrect bounds for my double integral. My bounds with respect to dy are
-sqrt(x)<= y >=sqrt(x)

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