- #1
kostoglotov
- 234
- 6
Homework Statement
Find the volume of the solid.
Under the paraboloid z = x^2 + y^2 and above the region bounded by y = x^2 and x = y^2
Well, those curves only intersects in the xy-plane at (0,0) and (1,1), and in the first Quadrant, and in that first Quadrant y = sqrt(x), and over that interval sqrt(x) >= x^2
So here is my working.
The Attempt at a Solution
[tex] \int \int_D (x^2+y^2) dA = \int_0^1 \int_{x^2}^{\sqrt(x)} x^2+y^2 dy dx [/tex]
[tex] \int_{x^2}^{\sqrt(x)} x^2+y^2 dy = \left[ yx^2+\frac{y^3}{3} \right]_{x^2}^{\sqrt(x)} [/tex]
[tex] = x^{1/5} + \frac{x^{1/3}}{3} - x^4 - \frac{x^6}{3} [/tex]
then
[tex] \int_0^1 x^{1/5} + \frac{x^{1/3}}{3} - x^4 - \frac{x^6}{3} dx = \left[\frac{5}{6}x^{5/6} + \frac{x^{4/3}}{4} - \frac{x^5}{5} - \frac{x^7}{21}\right]_0^1[/tex]
[tex] = \frac{5}{6} + \frac{1}{4} - \frac{1}{5} - \frac{1}{21} = \frac{117}{140} [/tex]
But the textbook gives the answer as [tex] \frac{6}{35} [/tex]
I can't figure out where I have gone wrong.