Section 15.3 : Double Integrals over General Regions
9. Evaluate ∬D7x2+14ydA where D is the region bounded by x=2y2 and x=8 in the order given below.
- Integrate with respect to x first and then y.
- Integrate with respect to y first and then x.
Start Solution
Here’s a quick sketch of the region with the curves labeled for integration with respect to x first.

The limits for the integral for integration with respect to x first are then,
−2≤y≤22y2≤x≤8Plugging these limits into the integral is then,
∬D7x2+14ydA=∫2−2∫82y27x2+14ydxdy Show Step 2The x integration for this integral is,
∬D7x2+14ydA=∫2−2∫82y27x2+14ydxdy=∫2−2(73x3+14xy)|82y2dy=∫2−235843+112y−28y3−563y6dy Show Step 3Finally, the y integration is,
∬D7x2+14ydA=∫2−235843+112y−28y3−563y6dy=(35843y+56y2−7y4−83y7)|2−2=4096b Integrate with respect to y first and then x. Show All Steps Hide All Steps
Start Solution
Here’s a quick sketch of the region with the curves labeled for integration with respect to y first.

Note that in order to do y integration first we needed to solve the equation of the parabola for y so the top and bottom curve will have distinct equations in terms of x, which we need to integrate with respect to y first.
The limits for the integral for integration with respect to y first are then,
0≤x≤8−√12x≤y≤√12xPlugging these limits into the integral is then,
∬D7x2+14ydA=∫80∫√12x−√12x7x2+14ydydx Show Step 2The y integration for this integral is,
∬D7x2+14ydA=∫80∫√12x−√12x7x2+14ydydx=∫80(7x2y+7y2)|√12x−√12xdx=∫8014√2x52dx Show Step 3Finally, the x integration is,
∬D7x2+14ydA=∫8014√2x52dx=(4√2x72)|80=4096We got the same result as the first order of integration as we knew we would.