# Double Integrals: Where am I making a mistake?

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1. Apr 13, 2015

### kostoglotov

1. The problem statement, all variables and given/known data

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.

3. The attempt at a solution

$$\int \int_D (x^2+y^2) dA = \int_0^1 \int_{x^2}^{\sqrt(x)} x^2+y^2 dy dx$$

$$\int_{x^2}^{\sqrt(x)} x^2+y^2 dy = \left[ yx^2+\frac{y^3}{3} \right]_{x^2}^{\sqrt(x)}$$

$$= x^{1/5} + \frac{x^{1/3}}{3} - x^4 - \frac{x^6}{3}$$

then

$$\int_0^1 x^{1/5} + \frac{x^{1/3}}{3} - x^4 - \frac{x^6}{3} dx = \left[\frac{5}{6}x^{5/6} + \frac{x^{4/3}}{4} - \frac{x^5}{5} - \frac{x^7}{21}\right]_0^1$$

$$= \frac{5}{6} + \frac{1}{4} - \frac{1}{5} - \frac{1}{21} = \frac{117}{140}$$

But the textbook gives the answer as $$\frac{6}{35}$$

I can't figure out where I have gone wrong.

2. Apr 13, 2015

### HallsofIvy

I have no idea how you got this last .line. $yx^2+ \frac{y^3}{3}$ evaluated at $y= x^{1/2}$ is
$(x^{1/2}x^2+ \frac{(x^{1/2})^3}{3}= x^{5/2}+ \frac{x^{3/2}}{3}$,, NOT "$x^{1/5}+ \frac{x^{1/3}}{3}$.

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3. Apr 13, 2015

### kostoglotov

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Yep, I'm tired. I just realized what I did wrong after running it through MATLAB...I really should just take a break and come back to it, before hastily posting here. Sorry :P