Evaluate the following integral. $\int_0^1 \int_1^{x^2} 3x-y \ dy\ dx$

Solution

Note the order of integration: the inner integral has a limit in terms of $x$, and the outer integral is evaluated with respect to $x$.

$\begin{aligned} \int_0^1 \int_1^{x^2} &3x-y \ dy\ dx\ &= \int_0^1 \bigg[ 3xy-\dfrac{1}{2}y^2 \bigg]_1^{x^2}\ dx\ &= \int_0^1 3x(x^2)-\dfrac{1}{2}(x^2)^2\ dx\ &= \int_0^1 3x^3-\dfrac{1}{2}x^4\ dx\ &= \dfrac{3}{4}x^4 - \dfrac{1}{10}x^5 \bigg|_0^1 = \ans{\dfrac{13}{20}} \end{aligned}$

Evaluate $\displaystyle\iint_R xy^2\ dA$, where $R$ is the triangular region with vertices $(0,0)$, $(2,4)$, and $(6,0)$.

Solution

We could take vertical slices, but we'd have to split the region into two. Instead, we'll take horizontal slices (with $Δy$), so we'll integrate with respect to $x$ first: $\int A(y)\ dy = \iint f(x,y)\ dx\ dy$

The limits for $y$ are from $0$ to $4$, and the limits for $x$ are from the left-hand function to the right-hand one:

$\begin{aligned} \iint_R xy^2\ dA &= \int_0^4 \int_{y/2}^{-y+6} xy^2\ dx\ dy\ &= \int_0^4 \bigg[\dfrac{1}{2}x^2y^2\bigg]_{y/2}^{-y+6}\ dy\ &= \int_0^4 \dfrac{3}{8}y^2-12y+36\ dy\ &= \dfrac{1}{8}y^3-6y^2+36y \bigg|_0^4 = \ans{56} \end{aligned}$

Evaluate $\displaystyle\iint_R xy\ dA$, where $R$ is bounded by $y=2x+1$, $y=−2x+5$, and $x=0$.

Solution

Here we'll take vertical slices, so we'll integrate with respect to $y$ first:

$\begin{aligned} \iint_R xy\ dA &= \int_0^1 \int_{2x+1}^{-2x+5} xy\ dy\ dx\ &= \int_0^1 \bigg[\dfrac{1}{2}xy^2\bigg]_{2x+1}^{-2x+5}\ dx\ &= \int_0^1 -12x^2+12x\ dx\ &= -4x^3+6x^2 \bigg|_0^1 = \ans{2} \end{aligned}$

Reverse the order of integration and evaluate $\displaystyle\int_0^1 \displaystyle\int_{x^2}^1 x^3 \sin (y^3)\ dy\ dx$.

Solution

$\begin{aligned} \int_0^1 \int_0^{\sqrt{y}} x^3 \sin (y^3)\ dx\ dy &= \int_0^1 \dfrac{1}{4}y^2 \sin (y^3)\ dy\ &= \ans{-\dfrac{1}{12}\cos 1 + 12} \end{aligned}$

Find the volume of the solid under $f(x,y)=x$ bounded by $y=x^2$, $y=−x+12$, and $y=4x+12$.

Solution

Notice that the intersection points are $−2$ and $3$, but we need to split this into two regions, one from $−2$ to $0$, and the other from $0$ to $3$.

$\begin{aligned} \textrm{Volume } &= \int_{-2}^0 \int_{x^2}^{4x+12} x\ dy\ dx + \int_0^3 \int_{x^2}^{-x+12} x\ dy\ dx\ &= \int_{-2}^0 x(4x+12)-x(x^2)\ dx + \int_0^3 x(-x+12)-x(x^2)\ dx\ &= \int_{-2}^0 -x^3+4x^2+12x\ dx + \int_0^3 -x^3-x^2+12x\ dx\ &= \bigg[-\dfrac{1}{4}x^4+\dfrac{4}{3}x^3+6x^2 \bigg]_{-2}^0 + \bigg[-\dfrac{1}{4}x^4-\dfrac{1}{3}x^3+6x^2 \bigg]_0^3\ &= \ans{\dfrac{185}{12}} \end{aligned}$