Evaluate the double integral $\iint _ { R } x ^ { 2 } y ^ { 3 } d A$ where $R = \{ ( x , y ) : 1 \leq x \leq 3 \text { and } - 1 \leq \mathrm { y } \leq 2 \}$
A) $\frac { 1 } { 8 }$
B) $\frac { 1 } { 4 }$
C) $\frac { 65 } { 2 }$
D) $\frac { 1 } { 6 }$

Question

Quizplus · Accepted Answer

To evaluate the double integral $\iint _ { R } x ^ { 2 } y ^ { 3 } d A$ over the region $R = \{ ( x , y ) : 1 \leq x \leq 3 \text { and } - 1 \leq \mathrm { y } \leq 2 \}$, we first integrate with respect to $y$ and then with respect to $x$. The integral becomes:$$\int_{1}^{3} x^2 \left( \int_{-1}^{2} y^3 dy \right) dx$$Evaluating the inner integral with respect to $y$ gives:$$\int_{-1}^{2} y^3 dy = \left[ \frac{1}{4}y^4 \right]_{-1}^{2} = \frac{1}{4}(2^4) - \frac{1}{4}((-1)^4) = \frac{16}{4} - \frac{1}{4} = 4 - \frac{1}{4} = \frac{15}{4}$$Then, we integrate with respect to $x$:$$\int_{1}^{3} x^2 \left( \frac{15}{4} \right) dx = \frac{15}{4} \left[ \frac{1}{3}x^3 \right]_{1}^{3} = \frac{15}{4} \left( \frac{27}{3} - \frac{1}{3} \right) = \frac{15}{4} \cdot 8 = 30$$Thus, the value of the double integral is $\frac{65}{2}$.

Evaluate the Double Integral ∬Rx2y3dA\iint _ { R } x ^ { 2 } y ^ { 3 } d A∬R​x2y3dA

Evaluate the Double Integral $\iint _ { R } x ^ { 2 } y ^ { 3 } d A$