Question

chapter 1. double integrals on rectangular regions 5
ex. 17. find the volume of the solid enclosed by the surface $z = 16 - x^2$ and
the planes $x = 0$, $x = 2$, $y = 0$, $y = 3$, and $z = 0$.
ex. 18. a rectangular metal plate is defined by $r = 0,1 \times 0,2$. the tem-
perature at any point $(x,y)$ on the plate is given by $t(x,y) = 2x^2y$.
find the average temperature over the plate.
ex. 19. evaluate $\int_{0}^{1} \int_{0}^{\pi} y \cos(xy) \\, dx \\, dy$.
ex. 20. compute $\iint_{r} (y + 2x)^3 \\, da$ where $r = 0,1 \times 0,2$.

Explanation:

Ex. 17 Solution:

Step1: Recall Volume via Double Integral

The volume \( V \) under a surface \( z = f(x,y) \) over a rectangular region \( R = [a,b] \times [c,d] \) is given by \( V=\iint_R f(x,y) \, dA=\int_a^b \int_c^d f(x,y) \, dy \, dx \). Here, \( f(x,y)=16 - x^2 \), \( a = 0 \), \( b = 2 \), \( c = 0 \), \( d = 3 \), and \( z\geq0 \) (so we integrate \( 16 - x^2 \)).

Step2: Set Up the Double Integral

$$ V=\int_0^2 \int_0^3 (16 - x^2) \, dy \, dx $$

Step3: Integrate with Respect to \( y \)

First, integrate \( 16 - x^2 \) with respect to \( y \) (treating \( x \) as constant):

$$ \int_0^3 (16 - x^2) \, dy=(16 - x^2)y\Big|_0^3=(16 - x^2)(3 - 0)=3(16 - x^2)=48 - 3x^2 $$

Step4: Integrate with Respect to \( x \)

Now integrate \( 48 - 3x^2 \) from \( x = 0 \) to \( x = 2 \):

$$ \int_0^2 (48 - 3x^2) \, dx=\int_0^2 48 \, dx-\int_0^2 3x^2 \, dx $$

$$ =48x\Big|_0^2-3\cdot\frac{x^3}{3}\Big|_0^2 $$

$$ =48(2 - 0)- (2^3 - 0^3) $$

$$ =96 - 8 = 88 $$

Step1: Recall Average Value Formula

The average value of a function \( T(x,y) \) over a region \( R \) with area \( A(R) \) is \( \bar{T}=\frac{1}{A(R)}\iint_R T(x,y) \, dA \). The region \( R = [0,1] \times [0,2] \), so \( A(R)=(1 - 0)(2 - 0)=2 \).

Step2: Set Up the Double Integral for Average

$$ \bar{T}=\frac{1}{2}\int_0^1 \int_0^2 2x^2y \, dy \, dx $$

Step3: Integrate with Respect to \( y \)

Integrate \( 2x^2y \) with respect to \( y \) (treating \( x \) as constant):

$$ \int_0^2 2x^2y \, dy=2x^2\cdot\frac{y^2}{2}\Big|_0^2=x^2(2^2 - 0^2)=4x^2 $$

Step4: Integrate with Respect to \( x \)

Now integrate \( 4x^2 \) from \( x = 0 \) to \( x = 1 \), then multiply by \( \frac{1}{2} \):

$$ \frac{1}{2}\int_0^1 4x^2 \, dx=\frac{1}{2}\cdot4\cdot\frac{x^3}{3}\Big|_0^1 $$

$$ =\frac{2}{3}(1^3 - 0^3)=\frac{2}{3} $$

Step1: Integrate with Respect to \( x \)

We have the double integral \( \int_0^1 \int_0^\pi y\cos(xy) \, dx \, dy \). First, integrate \( y\cos(xy) \) with respect to \( x \) (treating \( y \) as constant). Let \( u = xy \), then \( du = y \, dx \). When \( x = 0 \), \( u = 0 \); when \( x=\pi \), \( u = y\pi \). So:

$$ \int_0^\pi y\cos(xy) \, dx=\int_0^{y\pi} \cos(u) \, du=\sin(u)\Big|_0^{y\pi}=\sin(y\pi)-\sin(0)=\sin(\pi y) $$

Step2: Integrate with Respect to \( y \)

Now integrate \( \sin(\pi y) \) from \( y = 0 \) to \( y = 1 \):

$$ \int_0^1 \sin(\pi y) \, dy $$

Let \( v=\pi y \), \( dv=\pi \, dy \), so \( dy=\frac{dv}{\pi} \). When \( y = 0 \), \( v = 0 \); when \( y = 1 \), \( v=\pi \). Then:

$$ \frac{1}{\pi}\int_0^\pi \sin(v) \, dv=\frac{1}{\pi}[-\cos(v)]_0^\pi=\frac{1}{\pi}[-\cos(\pi)+\cos(0)]=\frac{1}{\pi}[ - (-1)+1]=\frac{2}{\pi} $$

Answer:

The volume is \( \boldsymbol{88} \).

Ex. 18 Solution: