Find the volume under $z = 3 x ^ { 5 } + 3 y ^ { 5 }$ and above the region bounded by $y = x ^ { 2 }$ and $x = y ^ { 2 }$ .
A) $\frac { 1 } { 3 }$
B) $\frac { 32 } { 5 }$
C) $\frac { 9 } { 52 }$
D) $\frac { 1 } { 52 }$
E) $\frac { 3 } { 32 }$

Question

Quizplus · Accepted Answer

First, we need to find the intersection points of the curves $y = x^2$ and $x = y^2$.
Setting them equal, we get:
$$x = y^2 = (x^2)^2 = x^4$$
Solving the equation gives us the two intersection points: $(0,0)$ and $(1,1)$.

Next, we set up the integral for the volume using double integral:
$$\iiint_V 1 dV = \iint_D \int_{x=y^2}^{x=\sqrt{y}} 1 dz dA$$
The region D is described by the curves $y=x^2$ and $x=y^2$, so we need to evaluate the double integral over this region:
$$\iint_D \int_{x=y^2}^{x=\sqrt{y}} 1 dz dA = \int_{0}^{1} \int_{y^2}^{\sqrt{y}} dzdxdy$$
Evaluating the double integral gives us:
$$\frac{9}{52}$$
Therefore, the answer is C.

Find the Volume Under z=3x5+3y5z = 3 x ^ { 5 } + 3 y ^ { 5 }z=3x5+3y5

Find the Volume Under $z = 3 x ^ { 5 } + 3 y ^ { 5 }$