Evaluate the triple integral:
∫ ∫ ∫ E xy dV, where E is bounded by the parabolic cylinders y =x2 and x = y2 and the planes z = 0 and z = x+ y.
The Correct Answer and Explanation is :
The given triple integral is:
[
\int \int \int_E xy \, dV
]
where ( E ) is the region bounded by the parabolic cylinders ( y = x^2 ), ( x = y^2 ), and the planes ( z = 0 ), and ( z = x + y ). We need to evaluate the integral.
Step 1: Understand the boundaries of the region
We first analyze the boundaries of the region described by the parabolic cylinders.
- ( y = x^2 ) and ( x = y^2 ) describe curves in the ( xy )-plane.
- The curve ( y = x^2 ) is a parabola opening upwards.
- The curve ( x = y^2 ) is also a parabola, but it opens sideways. These two curves intersect when ( y = x^2 ) and ( x = y^2 ). Substituting ( y = x^2 ) into ( x = y^2 ), we get ( x = (x^2)^2 = x^4 ). Solving for ( x ), we get ( x = 0 ) or ( x = 1 ). So, the region of integration in the ( xy )-plane is bounded by ( x = 0 ) and ( x = 1 ), with ( y ) ranging between ( x^2 ) and ( \sqrt{x} ).
- The planes ( z = 0 ) and ( z = x + y ) bound the region in the ( z )-direction. Hence, for each point in the ( xy )-plane, ( z ) will range from 0 to ( x + y ).
Step 2: Set up the triple integral
From the analysis above, the bounds for the triple integral are:
[
\int_0^1 \int_{x^2}^{\sqrt{x}} \int_0^{x + y} xy \, dz \, dy \, dx
]
Step 3: Evaluate the integral
We evaluate the integral step by step:
- Inner integral with respect to ( z ):
[
\int_0^{x+y} xy \, dz = xy(x + y)
]
- Middle integral with respect to ( y ):
[
\int_{x^2}^{\sqrt{x}} xy(x + y) \, dy
]
We expand the integrand:
[
xy(x + y) = x^2 y + x y^2
]
Now integrate term by term:
[
\int_{x^2}^{\sqrt{x}} x^2 y \, dy = x^2 \left[ \frac{y^2}{2} \right]_{x^2}^{\sqrt{x}} = x^2 \left( \frac{x}{2} – \frac{x^4}{2} \right) = \frac{x^3}{2} – \frac{x^6}{2}
]
[
\int_{x^2}^{\sqrt{x}} x y^2 \, dy = x \left[ \frac{y^3}{3} \right]_{x^2}^{\sqrt{x}} = x \left( \frac{x^{3/2}}{3} – \frac{x^6}{3} \right) = \frac{x^{5/2}}{3} – \frac{x^7}{3}
]
Summing these integrals gives:
[
\frac{x^3}{2} – \frac{x^6}{2} + \frac{x^{5/2}}{3} – \frac{x^7}{3}
]
- Outer integral with respect to ( x ):
Now we integrate each term with respect to ( x ) from 0 to 1:
[
\int_0^1 \left( \frac{x^3}{2} – \frac{x^6}{2} + \frac{x^{5/2}}{3} – \frac{x^7}{3} \right) dx
]
The integrals are:
[
\int_0^1 \frac{x^3}{2} \, dx = \frac{1}{2} \cdot \frac{x^4}{4} \Big|_0^1 = \frac{1}{8}
]
[
\int_0^1 \frac{x^6}{2} \, dx = \frac{1}{2} \cdot \frac{x^7}{7} \Big|_0^1 = \frac{1}{14}
]
[
\int_0^1 \frac{x^{5/2}}{3} \, dx = \frac{1}{3} \cdot \frac{x^{7/2}}{7/2} \Big|_0^1 = \frac{2}{21}
]
[
\int_0^1 \frac{x^7}{3} \, dx = \frac{1}{3} \cdot \frac{x^8}{8} \Big|_0^1 = \frac{1}{24}
]
Step 4: Sum up the results
Summing these results gives the value of the triple integral:
[
\frac{1}{8} – \frac{1}{14} + \frac{2}{21} – \frac{1}{24} = \frac{147}{168} – \frac{12}{168} + \frac{16}{168} – \frac{7}{168} = \frac{144}{168} = \frac{6}{7}
]
Thus, the value of the triple integral is:
[
\boxed{\frac{6}{7}}
]